3.3: Span, Basis, and Dimension
- Page ID
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Given a set of vectors, one can generate a vector space by forming all linear combinations of that set of vectors. The span of the set of vectors \(\{\text{v}_1,\: \text{v}_2,\cdots , \text{v}_n\}\) is the vector space consisting of all linear combinations of \(\text{v}_1,\: \text{v}_2,\cdots , \text{v}_n\). We say that a set of vectors spans a vector space.
For example, the set of three-by-one column matrices given by
\[\left\{\left(\begin{array}{c}1\\0\\0\end{array}\right),\:\left(\begin{array}{c}0\\1\\0\end{array}\right),\:\left(\begin{array}{c}2\\3\\0\end{array}\right)\right\}\nonumber \]
spans the vector space of all three-by-one matrices with zero in the third row. This vector space is a vector subspace of all three-by-one matrices.
One doesn’t need all three of these vectors to span this vector subspace because any one of these vectors is linearly dependent on the other two. The smallest set of vectors needed to span a vector space forms a basis for that vector space. Here, given the set of vectors above, we can construct a basis for the vector subspace of all three-by-one matrices with zero in the third row by simply choosing two out of three vectors from the above spanning set. Three possible bases are given by
\[\left\{\left(\begin{array}{c}1\\0\\0\end{array}\right),\:\left(\begin{array}{c}0\\1\\0\end{array}\right)\right\},\quad\left\{\left(\begin{array}{c}1\\0\\0\end{array}\right),\:\left(\begin{array}{c}2\\3\\0\end{array}\right)\right\},\quad\left\{\left(\begin{array}{c}0\\1\\0\end{array}\right),\:\left(\begin{array}{c}2\\3\\0\end{array}\right)\right\}.\nonumber \]
Although all three combinations form a basis for the vector subspace, the first combination is usually preferred because this is an orthonormal basis. The vectors in this basis are mutually orthogonal and of unit norm.
The number of vectors in a basis gives the dimension of the vector space. Here, the dimension of the vector space of all three-by-one matrices with zero in the third row is two.
Find an orthonormal basis for the set of all three-by-one matrices where the first row is equal to the third row.
Solution
There are many different solutions to this example, but a rather simple orthonormal basis is given by
\[\left\{\left(\begin{array}{c}0\\1\\0\end{array}\right),\frac{\sqrt{2}}{2}\left(\begin{array}{c}1\\0\\1\end{array}\right)\right\}.\nonumber \]
Any other three-by-one matrix with first row equal to third row can be written as a linear combination of these two basis vectors, and the dimension of this vector space is also two.
Determine a basis for \(\mathbb{P}_2\), the vector space consisting of all polynomials of degree less than or equal to two.
Solution
Again, there are many possible choices for a basis, but perhaps the simplest one is given by
\[\left\{\begin{array}{ccc}1,&x,&x^2\end{array}\right\}.\nonumber \]
Clearly, any polynomial of degree less than or equal to two can be written as a linear combination of these basis vectors. The dimension of \(\mathbb{P}_2\) is three.
Determine a basis for the vector space given by the general solution of the differential equation \(d^2y/dx^2 + y = 0\).
Solution
The general solution is given by
\[y(x)=a\cos x+b\sin x,\nonumber \]
and a basis for this vector space are just the functions
\[\left\{\begin{array}{cc}\cos x,&\sin x\end{array}\right\}.\nonumber \]
The dimension of the vector space given by the general solution of the differential equation is two. This dimension is equal to the order of the highest derivative in the differential equation.