Complete a matrix using eigenvectors

Question

I want to complete a matrix using eigenvectors For example the matrix $$A= \begin{pmatrix} 1 & a \\ 1 & b \\ \end{pmatrix} $$ have the eigenvectors $u=(1,1)$ and $v=(-2,1)$

The answer matrix is $$A= \begin{pmatrix} 1 & 2 \\ 1 & 2 \\ \end{pmatrix} $$ But how I can show $a=2$ and $b=2 $?

I tried using Diagonalization $SMS^{-1}$ with arbitrary scalars and Characteristic Polynomial of A but I can't find $a$ and $b$.

Answer 1

You can go straight from the definition of eigenvector. We know

$$\begin{bmatrix} 1&a\\1&b \end{bmatrix} \begin{bmatrix} 1\\1 \end{bmatrix} = \alpha \begin{bmatrix} 1\\1 \end{bmatrix} $$ and

$$\begin{bmatrix} 1&a\\1&b \end{bmatrix} \begin{bmatrix} -2\\1 \end{bmatrix} = \beta \begin{bmatrix} -2\\1 \end{bmatrix} $$ for some $\alpha$ and $\beta$. When we multiply out we get $$\begin{bmatrix} 1+a\\1+b \end{bmatrix} = \begin{bmatrix} \alpha \\ \alpha \end{bmatrix}$$ and $$\begin{bmatrix} -2+a\\-2+b \end{bmatrix} = \begin{bmatrix} -2 \beta \\ \beta \end{bmatrix}.$$

These tell us $$1+a=1+b$$ and $$-2+a=4-2b.$$ The first gives us that $a=b$. The latter then becomes $-2+a=4-2a$ or $3a=6$, so both $a$ and $b$ are $2$.

Answer 2

Simply use eigen vectors/values definition. Solve this linear system of 4 equations for ($a,b,\lambda_1,\lambda_2$)

$ A \left[ {\begin{array}{cc} 1 \\ 1 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 1 \\ 1 \\ \end{array} } \right]\lambda_1 $

$ A \left[ {\begin{array}{cc} -2 \\ 1 \\ \end{array} } \right]= \left[ {\begin{array}{cc} -2 \\ 1 \\ \end{array} } \right] \lambda_2 $

Where $\lambda_1$ and $\lambda_2$ are eigenvalues for each eigenvector