5.3: Similarity

Objectives

1. Learn to interpret similar matrices geoemetrically.
2. Understand the relationship between the eigenvalues, eigenvectors, and characteristic polynomials of similar matrices.
3. Recipe: compute \(Ax\) in terms of \(B\), \(C\) for \(A=CBC^{−1}\).
4. Picture: the geometry of similar matrices.
5. Vocabulary word: similarity.

Some matrices are easy to understand. For instance, a diagonal matrix

\[D=\left(\begin{array}{cc}2&0\\0&1/2\end{array}\right)\nonumber\]

just scales the coordinates of a vector: \(D\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}2x\\ y/2\end{array}\right)\). The purpose of most of the rest of this chapter is to understand complicated-looking matrices by analyzing to what extent they “behave like” simple matrices. For instance, the matrix

\[A=\frac{1}{10}\left(\begin{array}{cc}11&6\\9&14\end{array}\right)\nonumber\]

has eigenvalues \(2\) and \(1/2\), with corresponding eigenvectors \(v_1=\left(\begin{array}{c}2/3\\1\end{array}\right)\) and \(v_{2}=\left(\begin{array}{c}−1\\1\end{array}\right)\). Notice that

\[\begin{aligned}D(xe_{1}+ye_{2})=&xDe_{1}+yDe_{2}=2xe_{1}−\frac{1}{2}ye_{2} \\ A(xv_{1}+yv_{2})&=xAv_{1}+yAv_{2}=2xv_{1}−\frac{1}{2}yv_{2}.\end{aligned}\]

Using \(v_{1},\: v_{2}\) instead of the usual coordinates makes \(A\) “behave” like a diagonal matrix.

Figure \(\PageIndex{1}\): The matrices \(A\) and \(D\) behave similarly. Click “multiply” to multiply the colored points by \(D\) on the left and \(A\) on the right. (We will see in Section 5.4 why the points follow hyperbolic paths.)

The other case of particular importance will be matrices that “behave” like a rotation matrix: indeed, this will be crucial for understanding Section 5.5 geometrically. See Note \(\PageIndex{3}\).

In this section, we study in detail the situation when two matrices behave similarly with respect to different coordinate systems. In Section 5.4 and Section 5.5, we will show how to use eigenvalues and eigenvectors to find a simpler matrix that behaves like a given matrix.

Similar Matrices

We begin with the algebraic definition of similarity.

Definition \(\PageIndex{1}\): Similar

Two \(n\times n\) matrices \(A\) and \(B\) are similar if there exists an invertible \(n\times n\) matrix \(C\) such that \(A=CBC^{−1}\).

Example \(\PageIndex{1}\)

The matrices

\[\left(\begin{array}{cc}-12&15\\-10&13\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}3&0\\0&-2\end{array}\right)\nonumber\]

are similar because

\[\left(\begin{array}{cc}-12&15\\-10&13\end{array}\right)=\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)\left(\begin{array}{cc}3&0\\0&-2\end{array}\right)\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)^{-1},\nonumber\]

as the reader can verify.

Example \(\PageIndex{2}\)

The matrices

\[\left(\begin{array}{cc}3&0\\0&-2\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\nonumber\]

are not similar. Indeed, the second matrix is the identity matrix \(I_{2}\), so if \(C\) is any invertible \(2\times 2\) matrix, then

\[CI_{2}C^{-1}=CC^{-1}=I_{2}\neq\left(\begin{array}{cc}3&0\\0&-2\end{array}\right).\nonumber\]

As in the above example, one can show that \(I_{n}\) is the only matrix that is similar to \(I_{n}\), and likewise for any scalar multiple of \(I_{n}\).

Note \(\PageIndex{1}\)

Similarity is unrelated to row equivalence. Any invertible matrix is row equivalent to \(I_{n}\), but \(I_{n}\) is the only matrix similar to \(I_{n}\). For instance,

\[\left(\begin{array}{cc}2&1\\0&2\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\nonumber\]

are row equivalent but not similar.

As suggested by its name, similarity is what is called an equivalence relation. This means that it satisfies the following properties.

Proposition \(\PageIndex{1}\)

Let \(A\), \(B\), and \(C\) be \(n\times n\) matrices.

1. Reflexivity: \(A\) is similar to itself.
2. Symmetry: if \(A\) is similar to \(B\), then \(B\) is similar to \(A\).
3. Transitivity: if \(A\) is similar to \(B\) and \(B\) is similar to \(C\), then \(A\) is similar to \(C\).

Proof

1. Taking \(C=I_{n}=I_{n}^{−1}\), we have \(A=I_{n}AI_{n}^{−1}\).
2. Suppose that \(A=CBC^{−1}\). Multiplying both sides on the left by \(C^{−1}\) and on the right by \(C\) gives \[C^{−1}AC=C^{−1}(CBC^{−1})C=B.\nonumber\] Since \((C^{−1})^{−1}=C\), we have \(B=C^{−1}A(C^{−1})^{−1}\), so that \(B\) is similar to \(A\).
3. Suppose that \(A=DBD^{−1}\) and \(B=ECE^{−1}\). Subsituting for \(B\) and remembering that \((DE)^{−1}=E^{−1}D^{−1}\), we have \[A=D(ECE^{−1})D^{−1}=(DE)C(DE)^{−1},\nonumber\] which shows that \(A\) is similar to \(C\).

Example \(\PageIndex{3}\)

The matrices

\[\left(\begin{array}{cc}-12&15\\-10&13\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}3&0\\0&-2\end{array}\right)\nonumber\]

are similar, as we saw in Example \(\PageIndex{1}\). Likewise, the matrices 

\[\left(\begin{array}{cc}-12&15\\-10&13\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}-12&5\\-30&13\end{array}\right)\nonumber\]

are similar because

\[\left(\begin{array}{cc}-12&5\\-30&13\end{array}\right)=\left(\begin{array}{cc}2&-1\\2&1\end{array}\right)\left(\begin{array}{cc}-12&15\\-10&13\end{array}\right)\left(\begin{array}{cc}2&-1\\2&1\end{array}\right)^{-1}.\nonumber\]

It follows that 

\[\left(\begin{array}{cc}-12&5\\-30&13\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}3&0\\0&-2\end{array}\right)\nonumber\]

are similar to each other.

We conclude with an observation about similarity and powers of matrices.

Fact \(\PageIndex{1}\)

Let \(A=CBC^{-1}\). Then for any \(n\geq 1\), we have

\[A^{n}=CB^{n}C^{-1}.\nonumber\]

Proof

First note that

\[A^{2}=AA=(CBC^{−1})(CBC^{−1})=CB(C^{−1}C)BC^{−1}=CBI_{n}BC^{−1}=CB^{2}C^{−1}.\nonumber\]

Next we have

\[A^{3}=A^{2}A=(CB^{2}C^{−1})(CBC^{−1})=CB^{2}(C^{−1}C)BC^{−1}=CB^{3}C^{−1}.\nonumber\]

The pattern is clear.

Example \(\PageIndex{4}\)

Compute \(A^{100}\), where

\[A=\left(\begin{array}{cc}5&13\\-2&-5\end{array}\right)=\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)^{-1}.\nonumber\]

Solution

By the fact \(\PageIndex{1}\), we have

\[A^{100}=\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)^{100}\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)^{-1}.\nonumber\]

The matrix \(\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\) is a counterclockwise rotation by \(90^{\circ}\). If we rotate by \(90^{\circ}\) four times, then we end up where we started. Hence rotating by \(90^{\circ}\) one hundred times is the identity transformation, so

\[A^{100}=\left(\begin{array}{cc}-2&3\\1&-1\end{array}\right)\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\left(\begin{array}{c}-2&3\\1&-1\end{array}\right)^{-1}=\left(\begin{array}{cc}1&0\\0&1\end{array}\right).\nonumber\]

Geometry of Similar Matrices

Similarity is a very interesting construction when viewed geometrically. We will see that, roughly, similar matrices do the same thing in different coordinate systems. The reader might want to review \(\mathcal{B}\)-coordinates and nonstandard coordinate grids in Section 2.8 before reading this subsection.

By conditions 4 and 5 of the invertible matrix theorem in Section 5.1, an \(n\times n\) matrix \(C\) is invertible if and only if its columns \(v_{1},\:v_{2},\cdots ,v_{n}\) form a basis for \(\mathbb{R}^{n}\). This means we can speak of the \(\mathcal{B}\)-coordinates of a vector in \(\mathbb{R}^{n}\), where \(\mathcal{B}\) is the basis of columns of \(C\). Recall that

\[ [x]_{\mathcal{B}}=\left(\begin{array}{c}c_1 \\ c_2 \\ \vdots \\ c_n\end{array}\right)\quad\text{means}\quad x=c_1v_1+c_2v_2+\cdots +c_nv_n=\left(\begin{array}{cccc}|&|&\quad&| \\ v_1&v_2&\cdots&v_n \\ |&|&\quad&|\end{array}\right)\left(\begin{array}{c}c_1\\c_2\\ \vdots \\ c_n\end{array}\right).\nonumber\]

Since \(C\) is the matrix with columns \(v_1,\:v_2,\cdots, v_n\), this says that \(x=C[x]_{\mathcal{B}}\). Multiplying both sides by \(C^{−1}\) gives \([x]_{\mathcal{B}}=C^{−1}x\). To summarize:

Note \(\PageIndex{2}\)

Let \(C\) be an invertible \(n\times n\) matrix with columns \(v_1,\:v_2,\cdots ,v_n\), and let \(\mathcal{B}=\{v_1,\:v_2,\cdots ,v_n\}\), a basis for \(\mathbb{R}^{n}\). Then for any \(x\) in \(\mathbb{R}^{n}\), we have

\[C[x]_{\mathcal{B}}=x\quad\text{and}\quad C^{-1}x=[x]_{\mathcal{B}}.\nonumber\]

This says that \(C\) changes from the \(\mathcal{B}\)-coordinates to the usual coordinates, and \(C^{−1}\) changes from the usual coordinates to the \(\mathcal{B}\)-coordinates.

Suppose that \(A=CBC^{−1}\). The above observation gives us another way of computing \(Ax\) for a vector \(x\) in \(\mathbb{R}^{n}\). Recall that \(CBC^{−1}x=C(B(C^{−1}x))\), so that multiplying \(CBC^{−1}\) by \(x\) means first multiplying by \(C^{−1}\), then by \(B\), then by \(C\). See Example 3.4.10 in Section 3.4.

Recipe: Computing \(Ax\) in terms of \(B\).

Suppose that \(A=CBC^{−1}\), where \(C\) is an invertible matrix with columns \(v_{1},\:v_{2},\cdots ,v_n\). Let \(\mathcal{B}=\{v_1,\:v_2,\cdots ,v_n\}\), a basis for \(\mathbb{R}^{n}\). Let \(x\) be a vector in \(\mathbb{R}^{n}\). To compute \(Ax\), one does the following:

1. Multiply \(x\) by \(C^{−1}\), which changes to the \(\mathcal{B}\)-coordinates: \([x]_{\mathcal{B}}=C^{−1}x\).
2. Multiply this by \(B\): \(B[x]_{\mathcal{B}}=BC^{−1}x\).
3. Interpreting this vector as a \(\mathcal{B}\)-coordinate vector, we multiply it by \(C\) to change back to the usual coordinates: \(Ax=CBC^{−1}x=CB[x]_{\mathcal{B}}\).

Figure \(\PageIndex{2}\)

To summarize: if \(A=CBC^{-1}\), then \(A\) and \(B\) do the same thing, only in different coordinate systems.

The following example is the heart of this section.

Example \(\PageIndex{5}\)

Consider the matrices

\[A=\left(\begin{array}{cc}1/2&3/2\\ 3/2&1/2\end{array}\right)\quad B=\left(\begin{array}{cc}2&0\\0&-1\end{array}\right)\quad C=\left(\begin{array}{cc}1&1\\1&-1\end{array}\right).\nonumber\]

One can verify that \(A=CBC^{−1}\): see Example 5.4.1 in Section 5.4. Let \(v_{1}=\left(\begin{array}{c}1\\1\end{array}\right)\) and \(v_2=\left(\begin{array}{c}1\\-1\end{array}\right)\), the columns of \(C\), and let \(\mathcal{B}=\{v_1,\:v_2\}\), a basis of \(\mathbb{R}^{2}\).

The matrix \(B\) is diagonal: it scales the \(x\)-direction by a factor of \(2\) and the \(y\)-direction by a factor of \(−1\).

Figure \(\PageIndex{3}\)

To compute \(Ax\), first we multiply by \(C^{−1}\) to find the \(\mathcal{B}\)-coordinates of \(x\), then we multiply by \(B\), then we multiply by \(C\) again. For instance, let \(\color{Green}{x}\color{black}{=\left(\begin{array}{c}0\\-2\end{array}\right)}\).

1. We see from the \(\mathcal{B}\)-coordinate grid below that \(\color{Green}{x}\color{black}{=−}\color{Purple}{v_{1}}\color{black}{+}\color{blue}{v_{2}}\). Therefore, \(C^{−1}\color{Green}{x}\color{black}{=}\color{Green}{[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}-1\\1\end{array}\right)}\)..
2. Multiplying by \(B\) scales the coordinates: \(\color{red}{B[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}-2\\-1\end{array}\right)}\).
3. Interpreting \(\left(\begin{array}{c}-2\\-1\end{array}\right)\) as a \(\mathcal{B}\)-coordinate vector, we multiply by \(C\) to get

\[\color{red}{Ax}\color{black}{=C\left(\begin{array}{c}-2\\-1\end{array}\right)=-2}\color{Purple}{v_{1}}\color{black}{-}\color{blue}{v_{2}}\color{black}{=\left(\begin{array}{c}-3\\-1\end{array}\right).}\nonumber\]

Of course, this vector lies at \((−2,−1)\) on the \(\mathcal{B}\)-coordinate grid.

Figure \(\PageIndex{4}\)

Now let \(\color{Green}{x}\color{black}{=\frac{1}{2}\left(\begin{array}{c}5\\-3\end{array}\right)}\).

1. We see from the \(\mathcal{B}\)-coordinate grid that \(\color{Green}{x}\color{black}{=\frac{1}{2}}\color{Purple}{v_{1}}\color{black}{+2}\color{blue}{v_{2}}\). Therefore, \(C^{−1}\color{Green}{x}\color{black}{=}\color{Green}{[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}1/2 \\ 2\end{array}\right)}\).
2. Multiplying by \(B\) scales the coordinates: \(\color{Red}{B[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}1\\-2\end{array}\right)}\).
3. Interpreting \(\left(\begin{array}{c}1\\-2\end{array}\right)\) as a \(\mathcal{B}\)-coordinate vector, we multiply by \(C\) to get

\[\color{Red}{Ax}\color{black}{=C\left(\begin{array}{c}1\\-2\end{array}\right)=}\color{Purple}{v_{1}}\color{black}{-2}\color{blue}{v_{2}}\color{black}{=\left(\begin{array}{c}-1\\3\end{array}\right).}\nonumber\]

This vector lies at \((1,−2)\) on the \(\mathcal{B}\)-coordinate grid.

Figure \(\PageIndex{5}\)

To summarize:

• \(B\) scales the \(e_1\)-direction by \(2\) and the \(e_2\)-direction by \(-1\).
• \(A\) scales the \(v_1\)-direction by \(2\) and the \(v_2\)-direction by \(-1\).

Figure \(\PageIndex{6}\)

Figure \(\PageIndex{7}\): The geometric relationship between the similar matrices \(A\) and \(B\) acting on \(\mathbb{R}^2\). Click and drag the heads of \(x\) and \([x]_{\mathcal{B}}\). Study this picture until you can reliably predict where the other three vectors will be after moving one of them: this is the essence of the geometry of similar matrices.

Example \(\PageIndex{6}\): Interactive: Another matrix similar to \(B\)

Consider the matrices

\[A'=\frac{1}{5}\left(\begin{array}{cc}-8&-9\\6&13\end{array}\right)\quad B=\left(\begin{array}{cc}2&0\\0&-1\end{array}\right)\quad C'=\frac{1}{2}\left(\begin{array}{cc}-1&-3\\2&1\end{array}\right).\nonumber\]

Then \(A'=C'B(C')^{−1}\), as one can verify. Let \(v_1'=\frac{1}{2}\left(\begin{array}{c}-1\\2\end{array}\right)\) and \(v_2'=\frac{1}{2}\left(\begin{array}{c}-3\\1\end{array}\right)\), the columns of \(C'\), and let \(\mathcal{B}'=\{v_1',v_2'\}\). Then \(A'\) does the same thing as \(B\), as in the previous example \(\PageIndex{5}\), except \(A'\) uses the \(\mathcal{B}'\)-coordinate system. In other words:

• \(B\) scales the \(e_1\)-direction by \(2\) and the \(e_2\)-direction by \(-1\).
• \(A'\) scales the \(v_1'\)-direction by \(2\) and the \(v_2'\)-direction by \(-1\).

Figure \(\PageIndex{8}\)

Figure \(\PageIndex{9}\): The geometric relationship between the similar matrices \(A'\) and \(B\) acting on \(\mathbb{R}^2\). Click and drag the heads of \(x\) and \([x]_{\mathcal{B}'}\).

Example \(\PageIndex{7}\): A matrix similar to a rotation matrix

Consider the matrices

\[A=\frac{1}{6}\left(\begin{array}{cc}7&-17\\5&-7\end{array}\right)\quad B=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\quad C=\left(\begin{array}{cc}2&-1/2 \\ 1&1/2\end{array}\right).\nonumber\]

One can verify that \(A=CBC^{−1}\). Let \(v_{1}=\left(\begin{array}{c}2\\1\end{array}\right)\) and \(v_2=\frac{1}{2}\left(\begin{array}{c}-1\\1\end{array}\right)\), the columns of \(C\), and let \(\mathcal{B}=\{v_1,v_2\}\), a basis of \(\mathbb{R}^2\).

The matrix \(B\) rotates the plane counterclockwise by \(90^{\circ}\).

Figure \(\PageIndex{10}\)

To compute \(Ax\), first we multiply by \(C^{−1}\) to find the \(\mathcal{B}\)-coordinates of \(x\), then we multiply by \(B\), then we multiply by \(C\) again. For instance, let \(\color{Green}{x}\color{black}{=\frac{3}{2}\left(\begin{array}{c}1\\1\end{array}\right)}\).

1. We see from the \(\mathcal{B}\)-coordinate grid below that \(\color{Green}{x}\color{black}{=}\color{Purple}{v_{1}}\color{black}{+}\color{blue}{v_{2}}\). Therefore, \(C^{−1}\color{Green}{x}\color{black}{=}\color{Green}{[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}1\\1\end{array}\right)}\).
2. Multiplying by \(B\) rotates by \(90^{\circ}\): \(\color{Red}{B[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}-1\\1\end{array}\right)}\).
3. Interpreting \(\left(\begin{array}{c}-1\\1\end{array}\right)\) as a \(\mathcal{B}\)-coordinate vector, we multiply by \(C\) to get

\[\color{Red}{Ax}\color{black}{=C\left(\begin{array}{c}-1\\1\end{array}\right)=-}\color{Purple}{v_{1}}\color{black}{+}\color{blue}{v_{2}}\color{black}{=\frac{1}{2}\left(\begin{array}{c}-5\\1\end{array}\right).}\nonumber\]

Of course, this vector lies at \((-1,1)\) on the \(\mathcal{B}\)-coordinate grid.

Figure \(\PageIndex{11}\)

Now let \(\color{Green}{x}\color{black}{=\frac{1}{2}\left(\begin{array}{c}-1\\-2\end{array}\right)}\).

1. We see from the \(\mathcal{B}\)-coordinate grid that \(\color{Green}{x}\color{black}{=−\frac{1}{2}}\color{Purple}{v_{1}}\color{black}{−}\color{blue}{v_{2}}\). Therefore, \(C^{−1}\color{Green}{x}\color{black}{=}\color{Green}{[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}-1/2 \\ -1\end{array}\right)}\).
2. Multiplying by \(B\) rotates by \(90^{\circ}\): \(\color{Red}{B[x]_{\mathcal{B}}}\color{black}{=\left(\begin{array}{c}1\\ -1/2\end{array}\right)}\).
3. Interpreting \(\left(\begin{array}{c}1\\-1/2\end{array}\right)\) as a \(\mathcal{B}\)-coordinate vector, we multiply by \(C\) to get

\[\color{Red}{Ax}\color{black}{=C\left(\begin{array}{c}1\\-1/2\end{array}\right)=}\color{Purple}{v_{1}}\color{black}{-\frac{1}{2}}\color{blue}{v_{2}}\color{black}{=\frac{1}{4}\left(\begin{array}{c}9\\3\end{array}\right).}\nonumber\]

This vector lies at \((1,-\frac{1}{2})\) on the \(\mathcal{B}\)-coordinate grid.

Figure \(\PageIndex{12}\)

To summarize:

• \(B\) rotates counterclockwise around the circle centered at the origin and passing through \(e_1\) and \(e_2\).
• \(A\) rotates counterclockwise around the ellipse centered at the origin and passing through \(v_1\) and \(v_2\).

Figure \(\PageIndex{13}\)

Figure \(\PageIndex{14}\): The geometric relationship between the similar matrices \(A\) and \(B\) acting on \(\mathbb{R}^2\). Click and drag the heads of \(x\) and \([x]_{\mathcal{B}}\).

To summarize and generalize the previous example:

Note \(\PageIndex{3}\)

Let

\[B=\left(\begin{array}{cc}\cos\theta&-\sin\theta \\ \sin\theta&\cos\theta\end{array}\right)\quad C=\left(\begin{array}{cc}|&|\\v_1&v_2\\ |&|\end{array}\right)\quad A=CBC^{-1},\nonumber\]

where \(C\) is assumed invertible. Then:

• \(B\) rotates the plane by an angle of \(\theta\) around the circle centered at the origin and passing through \(e_1\) and \(e_2\), in the direction from \(e_1\) to \(e_2\).
• \(A\) rotates the plane by an angle of \(\theta\) around the ellipse centered at the origin and passing through \(v_1\) and \(v_2\), in the direction from \(v_1\) to \(v_2\).

Figure \(\PageIndex{15}\)

Example \(\PageIndex{8}\): Interactive: Similar \(3\times 3\) matrices

Consider the matrices

\[A=\left(\begin{array}{ccc}-1&0&0\\-1&0&2\\-1&1&1\end{array}\right)\quad B=\left(\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&2\end{array}\right)\quad C=\left(\begin{array}{ccc}-1&1&0\\1&1&1\\-1&0&1\end{array}\right).\nonumber\]

Then \(A=CBC^{−1}\), as one can verify. Let \(v_1,\: v_2,\: v_3\) be the columns of \(C\), and let \(\mathcal{B}=\{v_1,\: v_2,\: v_3\}\), a basis of \(\mathbb{R}^3\). Then \(A\) does the same thing as \(B\), except \(A\) uses the \(\mathcal{B}\)-coordinate system. In other words:

• \(B\) scales the \(e_1\), \(e_2\)-plane by \(-1\) and the \(e_3\)-direction by \(2\).
• \(A\) scales the \(v_1\), \(v_2\)-plane by \(-1\) and the \(v_3\)-direction by \(2\).

Figure \(\PageIndex{16}\): The geometric relationship between the similar matrices \(A\) and \(B\) acting on \(\mathbb{R}^3\). Click and drag the heads of \(x\) and \([x]_{\mathcal{B}}\).

Eigenvalues of Similar Matrices

Since similar matrices behave in the same way with respect to different coordinate systems, we should expect their eigenvalues and eigenvectors to be closely related.

Fact \(\PageIndex{2}\)

Similar matrices have the same characteristic polynomial.

Proof

Suppose that \(A=CBC^{−1}\), where \(A\), \(B\), \(C\) are \(n\times n\) matrices. We calculate

\[\begin{aligned}A-\lambda I_{n}&=CBC^{-1}-\lambda CC^{-1}=CBC^{-1}-C\lambda C^{-1} \\ &=CBC^{-1}-C\lambda I_{n}C^{-1}=C(B_\lambda I_{n})C^{-1}.\end{aligned}\]

Therefore,

\[\det(A-\lambda I_{n})=\det(C(B-\lambda I_{n})C^{-1})=\det(C)\det(B-\lambda I_{n})\det(C)^{-1}=\det(B-\lambda I_{n}).\nonumber\]

Here we have used the multiplicativity property Proposition 4.1.3 in Section 4.1 and its Corollary 4.1.2 in Section 4.1.

Since the eigenvalues of a matrix are the roots of its characteristic polynomial, we have shown:

Note \(\PageIndex{4}\)

Similar matrices have the same eigenvalues.

By Theorem 5.2.2 in Section 5.2, similar matrices also have the same trace and determinant.

Note \(\PageIndex{5}\)

The converse of fact \(\PageIndex{2}\) is false. Indeed, the matrices

\[\left(\begin{array}{cc}1&1\\0&1\end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\nonumber\]

both have the characteristic polynomial \(f(\lambda)=(\lambda -1)^{2}\), but they are not similar, because the only matrix that is similar to \(I_2\) is \(I_2\) itself.

Given that similar matrices have the same eigenvalues, one might guess that they have the same eigenvectors as well. Upon reflection, this is not what one should expect: indeed, the eigenvectors should only match up after changing from one coordinate system to another. This is the content of the next fact, remembering that \(C\) and \(C^{−1}\) change between the usual coordinates and the \(\mathcal{B}\)-coordinates.

Fact \(\PageIndex{3}\)

Suppose that \(A=CBC^{-1}\). Then

\[\begin{array}{lll}v\text{ is an eigenvector of }A&\implies& C^{-1}v\text{ is an eigenvector of }B \\ v\text{is an eigenvector of }B&\implies & Cv\text{ is an eigenvector of }A.\end{array}\nonumber\]

The eigenvalues of \(v/C^{-1}v\) or \(v/Cv\) are the same.

Proof

Suppose that \(v\) is an eigenvector of \(A\) with eigenvalue \(\lambda\), so that \(Av=\lambda v\). Then

\[B(C^{−1}v)=C^{−1}(CBC^{−1}v)=C^{−1}(Av)=C^{−1}\lambda v=\lambda (C^{−1}v),\nonumber\]

so that \(C^{−1}v\) is an eigenvector of \(B\) with eigenvalue \(\lambda\). Likewise if \(v\) is an eigenvector of \(B\) with eigenvalue \(\lambda\), then \(Bv=\lambda v\), and we have

\[A(Cv)=(CBC^{−1})Cv=CBv=C(\lambda v)=\lambda (Cv),\nonumber\]

so that \(Cv\) is an eigenvalue of \(A\) with eigenvalue \(\lambda\).

Note \(\PageIndex{6}\)

If \(A=CBC^{−1}\), then \(C^{−1}\) takes the \(\lambda\)-eigenspace of \(A\) to the \(\lambda\)-eigenspace of \(B\), and \(C\) takes the \(\lambda\)-eigenspace of \(B\) to the \(\lambda\)-eigenspace of \(A\).

Example \(\PageIndex{9}\)

We continue with the above example \(\PageIndex{5}\): let

\[A=\left(\begin{array}{cc}1/2&3/2\\3/2&1/2\end{array}\right)\quad B=\left(\begin{array}{cc}2&0\\0&-1\end{array}\right)\quad C=\left(\begin{array}{cc}1&1\\1&-1\end{array}\right),\nonumber\]

so \(A=CBC^{-1}\). Let \(v_1=\left(\begin{array}{c}1\\1\end{array}\right)\) and \(v_2=\left(\begin{array}{c}1\\-1\end{array}\right)\), the columns of \(C\). Recall that:

• \(B\) scales the \(e_1\)-direction by \(2\) and the \(e_2\)-direction by \(-1\).
• \(A\) scales the \(v_1\)-direction by \(2\) and the \(v_2\)-direction by \(-1\).

This means that the \(x\)-axis is the \(2\)-eigenspace of \(B\), and the \(y\)-axis is the \(−1\)-eigenspace of \(B\); likewise, the “\(v_1\)-axis” is the \(2\)-eigenspace of \(A\), and the “\(v_2\)-axis” is the \(−1\)-eigenspace of \(A\). This is consistent with the fact \(\PageIndex{3}\), as multiplication by \(C\) changes \(e_1\) into \(Ce_1=v_1\) and \(e_2\) into \(Ce_2=v_2\).

Figure \(\PageIndex{17}\)

Figure \(\PageIndex{18}\): The eigenspaces of \(A\) are the lines through \(v_1\) and \(v_2\), These are the images under \(C\) of the coordinate axes, which are the eigenspaces of \(B\).

Example \(\PageIndex{10}\): Interactive: Another matrix similar to \(B\)

Continuing with Example \(\PageIndex{6}\), let

\[A'=\frac{1}{5}\left(\begin{array}{cc}-8&-9\\6&13\end{array}\right)\quad B=\left(\begin{array}{cc}2&0\\0&-1\end{array}\right)\quad C'=\frac{1}{2}\left(\begin{array}{cc}-1&-3\\2&1\end{array}\right),\nonumber\]

so \(A'=C'B(C')^{−1}\). Let \(v_1'=\frac{1}{2}\left(\begin{array}{c}-1\\2\end{array}\right)\) and \(v_2'=\frac{1}{2}\left(\begin{array}{c}-3\\1\end{array}\right)\), the columns of \(C'\). Then:

• \(B\) scales the \(e_1\)-direction by \(2\) and the \(e_2\)-direction by \(-1\).
• \(A'\) scales the \(v_1'\)-direction by \(2\) and the \(v_2'\)-direction by \(-1\).

As before, the \(x\)-axis is the \(2\)-eigenspace of \(B\), and the \(y\)-axis is the \(−1\)-eigenspace of \(B\); likewise, the “\(v_1'\)-axis” is the \(2\)-eigenspace of \(A'\), and the “\(v_2'\)-axis” is the \(−1\)-eigenspace of \(A'\). This is consistent with fact \(\PageIndex{3}\), as multiplication by \(C'\) changes \(e_1\) into \(C'e_1=v_1'\) and \(e_2\) into \(C'e_2=v_2'\).

Figure \(\PageIndex{19}\)

Figure \(\PageIndex{20}\): The eigenspaces of \(A'\) are the lines through \(v_1'\) and \(v_2'\). These are the images under \(C'\) of the coordinate axes, which are the eigenspaces of \(B\).

Example \(\PageIndex{11}\): Interactive: Similar \(3\times 3\) matrices

Continuing with Example \(\PageIndex{8}\), let

\[A=\left(\begin{array}{ccc}-1&0&0\\-1&0&2\\-1&1&1\end{array}\right)\quad B=\left(\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&2\end{array}\right)\quad C=\left(\begin{array}{ccc}-1&1&0\\1&1&1\\-1&0&1\end{array}\right),\nonumber\]

so \(A=CBC^{-1}\). Let \(v_1\), \(v_2\), \(v_3\) by the columns of \(C\). Then:

• \(B\) scales the \(e_1\), \(e_2\)-plane by \(-1\) and the \(e_3\)-direction by \(2\).
• \(A\) scales the \(v_1\), \(v_2\)-plane by \(-1\) and the \(v_3\)-direction by \(2\).

In other words, the \(xy\)-plane is the \(−1\)-eigenspace of \(B\), and the \(z\)-axis is the \(2\)-eigenspace of \(B\); likewise, the “\(v_1\), \(v_2\)-plane” is the \(−1\)-eigenspace of \(A\), and the “\(v_3\)-axis” is the \(2\)-eigenspace of \(A\). This is consistent with fact \(\PageIndex{3}\), as multiplication by \(C\) changes \(e_1\) into \(Ce_1=v_1,\: e_2\) into \(Ce_2=v_2\), and \(e_3\) into \(Ce_3=v_3\).

Figure \(\PageIndex{21}\): The \(−1\)-eigenspace of \(A\) is the green plane, and the \(2\)-eigenspace of \(A\) is the violet line. These are the images under \(C\) of the \(xy\)-plane and the \(z\)-axis, respectively, which are the eigenspaces of \(B\).