• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q41E

Expert-verified Found in: Page 324 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find a basis of the linear space V of all ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrices A for which both $\left[\begin{array}{c}1\\ 1\end{array}\right]$ and $\left[\begin{array}{c}1\\ 2\end{array}\right]$ are eigenvectors, and thus determine the dimension of .

Hence, the required dimension is 2.

See the step by step solution

## Step 1: Definition of the Eigenvectors

Eigenvectors are a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector.

## Step 2: Find inverse of matrix

Let, S be a $2×2$ matrix whose columns are the vectors $\left[\begin{array}{c}1\\ 1\end{array}\right]$ and $\left[\begin{array}{c}1\\ 2\end{array}\right]$.

$S=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]$

Then, the matrix A can be calculated as,

S'AS=D

Here, D is the diagonal matrix.

Let it be: $D=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]$, where, a and b are the eigenvalues of A.

First, compute the inverse of the matrix S as follows:

${S}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2-1}\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]$

## Step 3: Finding dimension

Now, the matrix A can be written as,

$A=SD{S}^{1}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}2a& -a\\ -b& b\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2a-b& -a+b\\ 2a-2b& a+2b\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2a& -a\\ 2a& -a\end{array}\right]+\left[\begin{array}{cc}-b& b\\ -2b& 2b\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2& -1\\ 2& -1\end{array}\right]a+=\left[\begin{array}{cc}-1& 1\\ -2& 2\end{array}\right]b$

Thus, a basis of the linear space V is $\left\{\left[\begin{array}{cc}2& -1\\ 2& -1\end{array}\right],\left[\begin{array}{cc}-1& 1\\ -2& 2\end{array}\right]\right\}$, and from this it is clear that the dimension of V is dim V = 2. ### Want to see more solutions like these? 