# Finding Matrix A from Eigenvalues and Eigenvectors (Diagonalization)

Question:

Let $A$ be a $3 \times 3$ Matrix such that $[-3,4,1]$ is the eigenvector corresponding to eigenvalue $3$, and $[6,-3,2]$ is an eigenvector corresponding to the eigenvalue $2$. If $v$ = $[0,5,4]$, compute $A^{2}v$.

Hint: $[6,-3,2]+2[-3,4,1]$ = $v$

Where I'm stuck:

I know that we can solve A with the equation:

A = $P^{-1}DP$, where $P$ is the matrix consisting of the eigenvectors and $D$ is the matrix consisting of the eigenvalues.

However, since $A$ is a $3$ x $3$, we should have a $3 \times 3$ matrix $D$ and $3 \times 3$ matrix $P$ as well, consisting of the eigenvalues and eigenvectors respectively - and in this case, we only have two eigenvalues and two eigenvectors. Where can I obtain the other one of each?

Any help will be very appreciated.

Thank you.

• @Arjit: thank you very much for the edit! – Lucky Boy May 13 '16 at 21:08

You don't need to construct the matrix $A$ to answer this question. You just need to use the fact that $A$ is linear, i.e. that $A(\alpha u + \beta v) = \alpha Au + \beta A v$ for any vectors $u, v$ and scalars $\alpha, \beta$. Combine this with the information given about the eigenvectors and the hint and you can find your answer without knowing the matrix.
$$A^2(v)=A^2([6,-3,2]+2[-3,4,1])=A(A([6,-3,2]+2[-3,4,1]))=A(2[6,-3,2]+6[-3,4,1])=4[6,-3,2]+18[-3,4,1]$$
You want to find $A^2 v$, where $v=[0,5,4]$. You are not given $A$, but you know the image under $A$ of the two vectors $a=[-3,4,1]$ and $b=[6,-3,2]$. The first important point is that $A$ is a linear transformation, and so preserves sums, for example $A(a+b) = Aa+Ab$. The question is now whether $v$ can be expressed a linear combination of $a$ and $b$.
Observe that to get 0 in the first coordinate using a linear combination of $a$ and $b$, we must take $2a$ and $b$ and add them. If we do this, the other coordinates also add up, i.e. $v=2a+b$ (this is the second important point). Now use the fact that $Aa$ and $Ab$ are scalar multiples of $a$ and $b$, respectively (since they are eigenvectors).