# Determining a $4\times4$ matrix knowing $3$ of its $4$ eigenvectors and eigenvalues

The precise question given goes as follows;

The matrix 'A' $\in \Bbb R^{4 \times4}$ has eigenvectors $u_1, u_2, u_3, u_4$ where

$u_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \\ 1 \\ \end{pmatrix} , u_2 = \begin{pmatrix} 0 \\ 2 \\ 1 \\ -1 \\ \end{pmatrix}, u_3 = \begin{pmatrix} 3 \\ -1 \\ 1 \\ 2 \\ \end{pmatrix}$ satisfy:

$Au_1 = 2u_1,\; Au_2 = 14u_2,\; Au_3 = 18u_3$

Calculate A$w$ where $w = \begin{pmatrix} 49 \\ 13 \\ 47 \\ 18 \\ \end{pmatrix}$.

Usually I'd approach a question like this by using the relationship of the similar matrices $A = PA'P^{-1}$ where $A$ is the matrix , $P$ has the matrix $A$'s eigenvectors for columns, and $A'$ has the matrix $A$'s eigenvalues along its diagonal and $0$ elsewhere. And from there just do the calculation.

However only 3 of the 4 eigenvectors/values are given, so I dont know if this method is still applicable, and am having no success, or if I'm going down a rabbit hole and missing an obvious alternate.

Any suggestions? Thanks.

• Have you checked if $w$ is in $\operatorname{span}(u_1,u_2,u_3)$? – user491874 Jan 14 '18 at 12:24
• Do you know $A$? – Lord Shark the Unknown Jan 14 '18 at 12:29
• Let $U = [u_1 u_2 u_3]$ and $Uv = w$. Then $U^TUv = U^Tw$. Since $U^TU$ is invertible, we can find $v$ and verify that $w \in \mathrm{span}\{u_1,u_2,u_3\}$. – GNUSupporter 8964民主女神 地下教會 Jan 14 '18 at 12:33

The answer could be arbitrary unless $w$ is a combination of the given eigenvectors (if it's not then an arbitrary value of $Aw$ defines completely the linear map since you know the images of a linear basis).
One approach will be to find this combination -- it is $$w = 16 u_1 + 20 u_2 + 11 u_3.$$ From this you compute $A w$ easily.
Another approach is to complete $(u_1, u_2, u_3)$ by adding any (independent) fourth vector. This defines the matrix $P$ in your method. $A'$ is still diagonal, taking for fourth eigenvalue whatever you like since it will not change the result for $A w$ (though it does change $A$ of course).