Find matrix from Eigenvectors and Eigenvalues

Question

A matrix $A$ has eigenvectors
$v_1 = \left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)$ $v_2 = \left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)$

with corresponding eigenvalues $\lambda_1$= 2 and $\lambda_2$= -3, respectively.

Determine Ab for the vector b = $ \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)$

I know how to find eigenvalues and eigenvectors from a given matrix A, but not this one, the vector A is a 2x1 matrix, determinant does not exist here, so how to find the matrix A as stated in the question?

Answer 1

By definition of eigenvalue and eigenvector, we have $$\tag{1}A\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)=2\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)\mbox{ and }A\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)=-3\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right).$$ Now, since $$\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\frac{2}{3}\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)-\frac{1}{3}\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right),$$ we have $$A\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\frac{2}{3}A\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)-\frac{1}{3}A\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)=....\mbox{(using $(1)$)}$$

Answer 2

Let us try without computing $A$. To do that we have to decompose b as a linear combination of $v_1$ and $v_2$ like $\mathbf{b}=\alpha_1 v_1+\alpha_2 v_2$ And this would yield $$A\mathbf{b}=\alpha_1\lambda_1 v_1+\alpha_2\lambda_2 v_2$$ To find $\alpha_1$ and $\alpha_2$ we just have to solve a set of two linear equations $$\begin{cases} 2\alpha_1+\alpha_2=1\\ \alpha_1-\alpha_2=1 \end{cases}$$ (solution $\alpha_1=\frac{2}{3}$ and $\alpha_2=\frac{-1}{3}$)

Answer 3

Note that $$ \vec b=\frac{2}{3}\,\vec v_1-\frac{1}{3}\,\vec v_2 $$ Hence $$ A\vec b=\frac{2}{3}A\vec v_1-\frac{1}{3}A\vec v_2 $$

Answer 4

You don't have to know $A$ explicitly. All you need to know is that $Av_1=2v_1$ and $Av_2=-3v_2$. Since $b=\frac{2}{3}v_1-\frac{1}{3}v_2$ we have \begin{align*} Ab=A(2/3v_1-1/3v_2)=1/3(2Av_1-Av_2)=1/3(4v_1+3v_2)=1/3(11, 1). \end{align*}

But if you want to determine $A$, note that $TAT^{-1}=D$, where $D$ is the diagonal matrix of the eigenvalues of $A$, and the columns of $T$ are the corresponding eigenvectors of $A$.