2.2: Matrix Addition, Scalar Multiplication, and Transposition

A rectangular array of numbers is called a matrix (the plural is matrices), and the numbers are called the entries of the matrix. Matrices are usually denoted by uppercase letters: \(A\), \(B\), \(C\), and so on. Hence,

\[A = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 0 & 5 & 6 \end{array} \right] \quad B = \left[ \begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array} \right] \quad C = \left[ \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \right] \nonumber \]

are matrices. Clearly matrices come in various shapes depending on the number of rows and columns. For example, the matrix \(A\) shown has \(2\) rows and \(3\) columns. In general, a matrix with \(m\) rows and \(n\) columns is referred to as an \(\boldsymbol{m} \times \boldsymbol{n}\) matrix or as having size \(\boldsymbol{m} \times \boldsymbol{n}\). Thus matrices \(A\), \(B\), and \(C\) above have sizes \(2 \times 3\), \(2 \times 2\), and \(3 \times 1\), respectively. A matrix of size \(1 \times n\) is called a row matrix, whereas one of size \(m \times 1\) is called a column matrix. Matrices of size \(n \times n\) for some \(n\) are called square matrices.

Each entry of a matrix is identified by the row and column in which it lies. The rows are numbered from the top down, and the columns are numbered from left to right. Then the \(\boldsymbol{(i, j)}\)-entry of a matrix is the number lying simultaneously in row \(i\) and column \(j\). For example,

\[\begin{aligned} \mbox{The } (1, 2) \mbox{-entry of } &\left[ \begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right] \mbox{ is } -1. \\ \mbox{The } (2, 3) \mbox{-entry of } &\left[ \begin{array}{rrr} 1 & 2 & -1 \\ 0 & 5 & 6 \end{array}\right] \mbox{ is } 6.\end{aligned} \nonumber \]

A special notation is commonly used for the entries of a matrix. If \(A\) is an \(m \times n\) matrix, and if the \((i, j)\)-entry of \(A\) is denoted as \(a_{ij}\), then \(A\) is displayed as follows:

\[A = \left[ \begin{array}{ccccc} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn} \end{array} \right] \nonumber \]

This is usually denoted simply as \(A = \left[ a_{ij} \right]\). Thus \(a_{ij}\) is the entry in row \(i\) and column \(j\) of \(A\). For example, a \(3 \times 4\) matrix in this notation is written

\[A = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right] \nonumber \]

It is worth pointing out a convention regarding rows and columns: Rows are mentioned before columns. For example:

• If a matrix has size \(m \times n\), it has \(m\) rows and \(n\) columns.
• If we speak of the \((i, j)\)-entry of a matrix, it lies in row \(i\) and column \(j\).
• If an entry is denoted \(a_{ij}\), the first subscript \(i\) refers to the row and the second subscript \(j\) to the column in which \(a_{ij}\) lies.

Two points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) in the plane are equal if and only if¹ they have the same coordinates, that is \(x_{1} = x_{2}\) and \(y_{1} = y_{2}\). Similarly, two matrices \(A\) and \(B\) are called equal (written \(A = B\)) if and only if:

1. They have the same size.
2. Corresponding entries are equal.

If the entries of \(A\) and \(B\) are written in the form \(A = \left[ a_{ij} \right]\), \(B = \left[ b_{ij} \right]\), described earlier, then the second condition takes the following form:

\[A = \left[ a_{ij} \right] = \left[ b_{ij} \right] \mbox{ means } a_{ij} = b_{ij} \mbox{ for all } i \mbox{ and } j \nonumber \]

Example \(\PageIndex{1}\)

Given \(A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 3 & 0 & 1 \end{array} \right]\) and \(C = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) discuss the possibility that \(A = B\), \(B = C\), \(A = C\).

Solution

\(A = B\) is impossible because \(A\) and \(B\) are of different sizes: \(A\) is \(2 \times 2\) whereas \(B\) is \(2 \times 3\). Similarly, \(B = C\) is impossible. But \(A = C\) is possible provided that corresponding entries are equal: \(\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) means \(a = 1\), \(b = 0\), \(c = -1\), and \(d = 2\).

Matrix Addition

Definition: Matrix Addition \(\PageIndex{1}

If \(A\) and \(B\) are matrices of the same size, their sum \(A + B\) is the matrix formed by adding corresponding entries.

If \(A = \left[ a_{ij} \right]\) and \(B = \left[ b_{ij} \right]\), this takes the form

\[A + B = \left[ a_{ij} + b_{ij} \right] \nonumber \]

Note that addition is not defined for matrices of different sizes.

Example \(\PageIndex{2}\)

If \(A = \left[ \begin{array}{rrr} 2 & 1 & 3 \\ -1 & 2 & 0 \end{array} \right]\) and \(B = \left[ \begin{array}{rrr} 1 & 1 & -1 \\ 2 & 0 & 6 \end{array} \right]\), compute \(A + B\).

Solution

\[A + B = \left[ \begin{array}{rrr} 2 + 1 & 1 + 1 & 3 - 1 \\ -1 + 2 & 2 + 0 & 0 + 6 \end{array} \right] = \left[ \begin{array}{rrr} 3 & 2 & 2 \\ 1 & 2 & 6 \end{array} \right] \nonumber \]

Example \(\PageIndex{3}\)

Find \(a\), \(b\), and \(c\) if \(\left[ \begin{array}{ccc} a & b & c \end{array} \right] + \left[ \begin{array}{ccc} c & a & b \end{array} \right] = \left[ \begin{array}{ccc} 3 & 2 & -1 \end{array} \right]\).

Solution

Add the matrices on the left side to obtain

\[\left[ \begin{array}{ccc} a + c & b + a & c + b \end{array} \right] = \left[ \begin{array}{rrr} 3 & 2 & -1 \end{array} \right] \nonumber \]

Because corresponding entries must be equal, this gives three equations: \(a + c = 3\), \(b + a = 2\), and \(c + b = -1\). Solving these yields \(a = 3\), \(b = -1\), \(c = 0\).

If \(A\), \(B\), and \(C\) are any matrices of the same size, then

\[
\begin{array}{rlr}
A+B & =B+A &&&&& \text { (commutative law) } \\
A+(B+C) & =(A+B)+C &&&&& \text { (associative law) }
\end{array}
\]
(commutative law)
(associative law)

In fact, if \(A = \left[ a_{ij} \right]\) and \(B = \left[ b_{ij} \right]\), then the \((i, j)\)-entries of \(A + B\) and \(B + A\) are, respectively, \(a_{ij} + b_{ij}\) and \(b_{ij} + a_{ij}\). Since these are equal for all \(i\) and \(j\), we get

\[A + B = \left[ \begin{array}{c} a_{ij} + b_{ij} \end{array} \right] = \left[ \begin{array}{c} b_{ij} + a_{ij} \end{array} \right] = B + A \nonumber \]

The associative law is verified similarly.

The \(m \times n\) matrix in which every entry is zero is called the \(m \times n\) zero matrix and is denoted as \(0\) (or \(0_{mn}\) if it is important to emphasize the size). Hence,

\[0 + X = X \nonumber \]

holds for all \(m \times n\) matrices \(X\). The negative of an \(m \times n\) matrix \(A\) (written \(-A\)) is defined to be the \(m \times n\) matrix obtained by multiplying each entry of \(A\) by \(-1\). If \(A = \left[ a_{ij} \right]\), this becomes \(-A = \left[ -a_{ij} \right]\). Hence,

\[A + (-A) = 0 \nonumber \]

holds for all matrices \(A\) where, of course, \(0\) is the zero matrix of the same size as \(A\).

A closely related notion is that of subtracting matrices. If \(A\) and \(B\) are two \(m \times n\) matrices, their difference \(A - B\) is defined by

\[A - B = A + (-B) \nonumber \]

Note that if \(A = \left[ a_{ij} \right]\) and \(B = \left[ b_{ij} \right]\), then

\[A - B = \left[ a_{ij} \right] + \left[ -b_{ij} \right] = \left[ a_{ij} - b_{ij} \right] \nonumber \]

is the \(m \times n\) matrix formed by subtracting corresponding entries.

Example \(\PageIndex{4}\)

Let \(A = \left[ \begin{array}{rrr} 3 & -1 & 0 \\ 1 & 2 & -4 \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 1 & -1 & 1 \\ -2 & 0 & 6 \end{array} \right]\), \(C = \left[ \begin{array}{rrr} 1 & 0 & -2 \\ 3 & 1 & 1 \end{array} \right]\). Compute \(-A\), \(A - B\), and
\(A + B - C\).

Solution

\[\begin{aligned} -A &= \left[ \begin{array}{rrr} -3 & 1 & 0 \\ -1 & -2 & 4 \end{array} \right] \\ A - B &= \left[ \begin{array}{lcr} 3 - 1 & -1 - (-1) & 0 - 1 \\ 1 - (-2) & 2 - 0 & -4 - 6 \end{array} \right] = \left[ \begin{array}{rrr} 2 & 0 & -1 \\ 3 & 2 & -10 \end{array} \right] \\ A + B - C &= \left[ \begin{array}{rrl} 3 + 1 - 1 & -1 - 1 - 0 & 0 + 1 -(-2) \\ 1 - 2 - 3 & 2 + 0 - 1 & -4 + 6 -1 \end{array} \right] = \left[ \begin{array}{rrr} 3 & -2 & 3 \\ -4 & 1 & 1 \end{array} \right]\end{aligned} \nonumber \]

Example \(\PageIndex{5}\)

Solve \(\left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right] + X = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) where \(X\) is a matrix.

Solution

We solve a numerical equation \(a + x = b\) by subtracting the number \(a\) from both sides to obtain \(x = b - a\). This also works for matrices. To solve \(\left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right] + X = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right]\) simply subtract the matrix \(\left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right]\) from both sides to get

\[X = \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right] - \left[ \begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array} \right] = \left[ \begin{array}{cr} 1 - 3 & 0 - 2 \\ -1 - (-1) & 2 - 1 \end{array} \right] = \left[ \begin{array}{rr} -2 & -2 \\ 0 & 1 \end{array} \right] \nonumber \]

The reader should verify that this matrix \(X\) does indeed satisfy the original equation.

The solution in Example \(\PageIndex{5}\) solves the single matrix equation \(A + X = B\) directly via matrix subtraction: \(X = B - A\). This ability to work with matrices as entities lies at the heart of matrix algebra.

It is important to note that the sizes of matrices involved in some calculations are often determined by the context. For example, if

\[A + C = \left[ \begin{array}{rrr} 1 & 3 & -1 \\ 2 & 0 & 1 \end{array} \right] \nonumber \]

then \(A\) and \(C\) must be the same size (so that \(A + C\) makes sense), and that size must be \(2 \times 3\) (so that the sum is \(2 \times 3\)). For simplicity we shall often omit reference to such facts when they are clear from the context.

Scalar Multiplication

In gaussian elimination, multiplying a row of a matrix by a number \(k\) means multiplying every entry of that row by \(k\).

Definition: Matrix Scalar Multiplication \(PageIndex{2}\)

More generally, if \(A\) is any matrix and \(k\) is any number, the scalar multiple \(kA\) is the matrix obtained from \(A\) by multiplying each entry of \(A\) by \(k\).

If \(A = \left[ a_{ij} \right]\), this is

\[kA = \left[ ka_{ij} \right] \nonumber \]

Thus \(1A = A\) and \((-1)A = -A\) for any matrix \(A\).

The term scalar arises here because the set of numbers from which the entries are drawn is usually referred to as the set of scalars. We have been using real numbers as scalars, but we could equally well have been using complex numbers.

Example \(\PageIndex{6}\)

If \(A = \left[ \begin{array}{rrr} 3 & -1 & 4 \\ 2 & 0 & 6 \end{array} \right]\) and \(B = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 0 & 3 & 2 \end{array} \right]\) compute \(5A\), \(\frac{1}{2}B\), and \(3A - 2B\).

Solution

\[\begin{aligned} 5A &= \left[ \begin{array}{rrr} 15 & -5 & 20 \\ 10 & 0 & 30 \end{array} \right], \quad \frac{1}{2}B = \left[ \begin{array}{rrr} \frac{1}{2} & 1 & -\frac{1}{2} \\ 0 & \frac{3}{2} & 1 \end{array} \right] \\ 3A - 2B &= \left[ \begin{array}{rrr} 9 & -3 & 12 \\ 6 & 0 & 18 \end{array} \right] - \left[ \begin{array}{rrr} 2 & 4 & -2 \\ 0 & 6 & 4 \end{array} \right] = \left[ \begin{array}{rrr} 7 & -7 & 14 \\ 6 & -6 & 14 \end{array} \right]\end{aligned} \nonumber \]

If \(A\) is any matrix, note that \(kA\) is the same size as \(A\) for all scalars \(k\). We also have

\[0A = 0 \quad \mbox{ and } \quad k0 = 0 \nonumber \]

because the zero matrix has every entry zero. In other words, \(kA = 0\) if either \(k = 0\) or \(A = 0\). The converse of this statement is also true, as Example [exa:002159] shows.

Example \(\PageIndex{7}\)

If \(kA = 0\), show that either \(k = 0\) or \(A = 0\).

Solution

Write \(A = \left[ a_{ij} \right]\) so that \(kA = 0\) means \(ka_{ij} = 0\) for all \(i\) and \(j\). If \(k = 0\), there is nothing to do. If \(k \neq 0\), then \(ka_{ij} = 0\) implies that \(a_{ij} = 0\) for all \(i\) and \(j\); that is, \(A = 0\).

For future reference, the basic properties of matrix addition and scalar multiplication are listed in Theorem \(\PageIndex{1}\).

Theorem \(\PageIndex{1}\)

Let \(A\), \(B\), and \(C\) denote arbitrary \(m \times n\) matrices where \(m\) and \(n\) are fixed. Let \(k\) and \(p\) denote arbitrary real numbers. Then

1. \(A + B = B + A\).
2. \(A + (B + C) = (A + B) + C\).
3. There is an \(m \times n\) matrix \(0\), such that \(0 + A = A\) for each \(A\).
4. For each \(A\) there is an \(m \times n\) matrix, \(-A\), such that \(A + (-A) = 0\).
5. \(k(A + B) = kA + kB\).
6. \((k + p)A = kA + pA\).
7. \((kp)A = k(pA)\).
8. \(1A = A\).

Proof. Properties 1–4 were given previously. To check Property 5, let \(A = \left[ a_{ij} \right]\) and \(B = \left[ b_{ij} \right]\) denote matrices of the same size. Then \(A + B = \left[ a_{ij} + b_{ij} \right]\), as before, so the \((i, j)\)-entry of \(k(A + B)\) is

\[k(a_{ij} + b_{ij}) = ka_{ij} + kb_{ij} \nonumber \]

But this is just the \((i, j)\)-entry of \(kA + kB\), and it follows that \(k(A + B) = kA + kB\). The other Properties can be similarly verified; the details are left to the reader.

The Properties in Theorem [thm:002170] enable us to do calculations with matrices in much the same way that numerical calculations are carried out. To begin, Property 2 implies that the sum

\[(A + B) + C = A + (B + C) \nonumber \]

is the same no matter how it is formed and so is written as \(A + B + C\). Similarly, the sum

\[A + B + C + D \nonumber \]

is independent of how it is formed; for example, it equals both \((A + B) + (C + D)\) and \(A + \left[ B + (C + D) \right]\). Furthermore, property 1 ensures that, for example,

\[B + D + A + C = A + B + C + D \nonumber \]

In other words, the order in which the matrices are added does not matter. A similar remark applies to sums of five (or more) matrices.

Properties 5 and 6 in Theorem [thm:002170] are called distributive laws for scalar multiplication, and they extend to sums of more than two terms. For example,

\[k(A + B - C) = kA + kB - kC \nonumber \]

\[(k + p - m)A = kA + pA -mA \nonumber \]

Similar observations hold for more than three summands. These facts, together with properties 7 and 8, enable us to simplify expressions by collecting like terms, expanding, and taking common factors in exactly the same way that algebraic expressions involving variables and real numbers are manipulated. The following example illustrates these techniques.

Example \(\PageIndex{8}\)

Simplify \(2(A + 3C) - 3(2C - B) - 3 \left[ 2(2A + B - 4C) - 4(A - 2C) \right]\) where \(A\), \(B\), and \(C\) are all matrices of the same size.

Solution

The reduction proceeds as though \(A\), \(B\), and \(C\) were variables.

\[\begin{aligned} 2(A &+ 3C) - 3(2C - B) - 3 \left[ 2(2A + B - 4C) - 4(A - 2C) \right] \\ &= 2A + 6C - 6C + 3B - 3 \left[ 4A + 2B - 8C - 4A + 8C \right] \\ &= 2A + 3B - 3 \left[ 2B \right] \\ &= 2A - 3B\end{aligned} \nonumber \]

Transpose of a Matrix

Many results about a matrix \(A\) involve the rows of \(A\), and the corresponding result for columns is derived in an analogous way, essentially by replacing the word row by the word column throughout. The following definition is made with such applications in mind.

Definition: Transpose of a Matrix \(\PageIndex{3}\)

If \(A\) is an \(m \times n\) matrix, the transpose of \(A\), written \(A^{T}\), is the \(n \times m\) matrix whose rows are just the columns of \(A\) in the same order.

In other words, the first row of \(A^{T}\) is the first column of \(A\) (that is it consists of the entries of column 1 in order). Similarly the second row of \(A^{T}\) is the second column of \(A\), and so on.

Example \(\PageIndex{9}\)

Write down the transpose of each of the following matrices.

\[A = \left[ \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \right] \quad B = \left[ \begin{array}{rrr} 5 & 2 & 6 \end{array} \right] \quad C = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array} \right] \quad D = \left[ \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & 2 \\ -1 & 2 & 1 \end{array} \right] \nonumber \]

Solution

\[A^{T} = \left[ \begin{array}{rrr} 1 & 3 & 2 \end{array} \right],\ B^{T} = \left[ \begin{array}{r} 5 \\ 2 \\ 6 \end{array} \right],\ C^{T} = \left[ \begin{array}{rrr} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array} \right], \mbox{ and } D^{T} = D. \nonumber \]

If \(A = \left[ a_{ij} \right]\) is a matrix, write \(A^{T} = \left[ b_{ij} \right]\). Then \(b_{ij}\) is the \(j\)th element of the \(i\)th row of \(A^{T}\) and so is the \(j\)th element of the \(i\)th column of \(A\). This means \(b_{ij} = a_{ji}\), so the definition of \(A^{T}\) can be stated as follows:

\[\label{eq:transpose} \mbox{If } A = \left[ a_{ij} \right] \mbox{, then } A^{T} = \left[ a_{ji} \right]. \]

This is useful in verifying the following properties of transposition.

Theorem \(\PageIndex{1}\)

Let \(A\) and \(B\) denote matrices of the same size, and let \(k\) denote a scalar.

1. If \(A\) is an \(m \times n\) matrix, then \(A^{T}\) is an \(n \times m\) matrix.
2. \((A^{T})^{T} = A\).
3. \((kA)^{T} = kA^{T}\).
4. \((A + B)^{T} = A^{T} + B^{T}\).

Proof. Property 1 is part of the definition of \(A^{T}\), and Property 2 follows from ([eq:transpose]). As to Property 3: If \(A = \left[ a_{ij} \right]\), then \(kA = \left[ ka_{ij} \right]\), so ([eq:transpose]) gives

\[(kA)^{T} = \left[ ka_{ji} \right] = k \left[ a_{ji} \right] = kA^{T} \nonumber \]

Finally, if \(B = \left[ b_{ij} \right]\), then \(A + B = \left[ c_{ij} \right]\) where \(c_{ij} = a_{ij} + b_{ij}\) Then ([eq:transpose]) gives Property 4:

\[(A + B)^{T} = \left[ c_{ij} \right]^{T} = \left[ c_{ji} \right] = \left[ a_{ji} + b_{ji} \right] = \left[ a_{ji} \right] + \left[ b_{ji} \right] = A^{T} + B^{T} \nonumber \]

There is another useful way to think of transposition. If \(A = \left[ a_{ij} \right]\) is an \(m \times n\) matrix, the elements \(a_{11}, a_{22}, a_{33}, \dots\) are called the main diagonal of \(A\). Hence the main diagonal extends down and to the right from the upper left corner of the matrix \(A\); shown in bold in the following examples:

\[ \begin{pmatrix}
\mathbf{a_{11}} & a_{12} \\
a_{21} & \mathbf{ a_{22} } \\
a_{31} & a_{32}
\end{pmatrix} \begin{pmatrix}
\mathbf{a_{11}} & a_{12} & a_{13} \\
a_{21} & \mathbf{ a_{22}} & a_{23}
\end{pmatrix} \begin{pmatrix}
\mathbf{a_{11}} & a_{12} & a_{13} \\
a_{21} & \mathbf{a_{22}} & a_{23} \\
a_{31} & a_{32} & \mathbf{ a_{33}}
\end{pmatrix} \begin{pmatrix}
\mathbf{ a_{11}} \\
a_{21}
\end{pmatrix}\]

Thus forming the transpose of a matrix \(A\) can be viewed as “flipping” \(A\) about its main diagonal, or as “rotating” \(A\) through \(180^{\circ}\) about the line containing the main diagonal. This makes Property 2 in Theorem \(\PageIndex{2}\) transparent.

Example \(\PageIndex{10}\)

Solve for \(A\) if \(\left(2A^{T} - 3 \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right] \right)^{T} = \left[ \begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array} \right]\).

Solution

Using Theorem \(\PageIndex{2}\), the left side of the equation is

\[\left(2A^{T} - 3 \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right]\right)^{T} = 2\left(A^{T}\right)^{T} - 3 \left[ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right]^{T} = 2A - 3 \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] \nonumber \]

Hence the equation becomes

\[2A - 3 \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] = \left[ \begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array} \right] \nonumber \]

Thus \(2A = \left[ \begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array} \right] + 3 \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right] = \left[ \begin{array}{rr} 5 & 0 \\ 5 & 5 \end{array} \right]\), so finally \(A = \frac{1}{2} \left[ \begin{array}{rr} 5 & 0 \\ 5 & 5 \end{array} \right] = \frac{5}{2} \left[ \begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right]\).

Note that Example [exa:002281] can also be solved by first transposing both sides, then solving for \(A^{T}\), and so obtaining \(A = (A^{T})^{T}\). The reader should do this.

The matrix \(D = \left[ \begin{array}{rr} 1 & 2 \\ 2 & 5 \end{array}\right]\) in Example [exa:002220] has the property that \(D = D^{T}\). Such matrices are important; a matrix \(A\) is called symmetric if \(A = A^{T}\). A symmetric matrix \(A\) is necessarily square (if \(A\) is \(m \times n\), then \(A^{T}\) is \(n \times m\), so \(A = A^{T}\) forces \(n = m\)). The name comes from the fact that these matrices exhibit a symmetry about the main diagonal. That is, entries that are directly across the main diagonal from each other are equal.

For example, \(\left[ \begin{array}{ccc} a & b & c \\ b^\prime & d & e \\ c^\prime & e^\prime & f \end{array} \right]\) is symmetric when \(b = b^\prime\), \(c = c^\prime\), and \(e = e^\prime\).

Example \(\PageIndex{11}\)

If \(A\) and \(B\) are symmetric \(n \times n\) matrices, show that \(A + B\) is symmetric.

Solution

We have \(A^{T} = A\) and \(B^{T} = B\), so, by Theorem [thm:002240], we have \((A + B)^{T} = A^{T} + B^{T} = A + B\). Hence \(A + B\) is symmetric.

Example \(\PageIndex{12}\)

Suppose a square matrix \(A\) satisfies \(A = 2A^{T}\). Show that necessarily \(A = 0\).

Solution

If we iterate the given equation, Theorem [thm:002240] gives

\[A = 2A^{T} = 2 {\left[ 2A^{T} \right]}^T = 2 \left[ 2(A^{T})^{T} \right] = 4A \nonumber \]

Subtracting \(A\) from both sides gives \(3A = 0\), so \(A = \frac{1}{3}(0) = 0\).

---

1. If \(p\) and \(q\) are statements, we say that \(p\) implies \(q\) if \(q\) is true whenever \(p\) is true. Then “\(p\) if and only if \(q\)” means that both \(p\) implies \(q\) and \(q\) implies \(p\). See Appendix [chap:appbproofs] for more on this.↩