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Let A be a 4x4 matrix with eigenvalues $\lambda$ = 2,3 and eigenspaces

$E_{\lambda=2} = \operatorname{span} \left\{ {\begin{bmatrix} 1\\0\\0\\1\end{bmatrix}, \begin{bmatrix} 1\\0\\1\\0\end{bmatrix}}\right\}$ and $E_{\lambda=3} = \operatorname{span} \left\{ {\begin{bmatrix} 0\\1\\0\\1\end{bmatrix}, \begin{bmatrix} 0\\0\\1\\1\end{bmatrix}}\right\}$

(A) Can you calculate the algebraic multiplicities of $\lambda$ = 2 and $\lambda$ = 3? If yes, do it.
(B) Can you calculate A? If yes, how?

I know how to calculate the algebraic multiplicities given the matrix A, but I do not know where to start when I am only given the eigenvalues and eigenspaces. My guess as to how to find A is to use the theorem A $\vec x$ = $\lambda$$\vec x$
For example, later in the problem I am given $\vec x = \begin{bmatrix} 3\\1\\4\\6\end{bmatrix} $
Solving A $\vec x$ = $\lambda$ $\vec x$, I get $A\begin{bmatrix} 3\\1\\4\\6\end{bmatrix} = \begin{bmatrix} 6\\3\\11\\16\end{bmatrix}$

How do I find A from here?

Additionally, I know that I cannot recover A from the eigenpairs if A is deficient, so I know I need to solve (a) first in order to solve (b). However, I have no idea how to find the algebraic multiplicities of A without first knowing A and finding the characteristic polynomial.

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  • $\begingroup$ Get us started, Sarah... $\endgroup$ – JohnD Dec 9 '14 at 22:23
  • $\begingroup$ What are your thoughts on the problem? Do you think that either can be calculated, any thoughts as to how? Did you have any other ideas? $\endgroup$ – Omnomnomnom Dec 9 '14 at 22:23
  • $\begingroup$ Essential: Down with Determinants! (axler.net/DwD.html) or why the algebraic multiplicity is as geometric as the geometric multiplicity. $\endgroup$ – Martín-Blas Pérez Pinilla Dec 11 '14 at 8:20
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A Long Hint: (thanks for your compliance and your patience!)

Recall that the geometric multiplicities are always at most the algebraic multiplicities; in other words $A(\lambda)\geq G(\lambda)$. By the information we already have, we know that $A(2)\geq 2$ and $A(3)\geq 2$.

Moreover, some general theory: we know that the characteristic polynomial has degree $n$, and by the Fundamental Theorem of Algebra every degree-$n$ real polynomial has $n$ complex roots. Therefore, $\sum_\lambda A(\lambda)=n$.

Applying the general theory to our situation, we have $A(2)+A(3)=4$.

Can you see how to proceed?

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  • $\begingroup$ Yes thank you! I knew A(2) + A(3) had to equal 4, but I wasn't sure how to tell if A(2) = 1 or 3, and vice versa. So thank you for the comparison to the geometric multiplicity. This makes sense now. Thank you for your help! $\endgroup$ – Sarah Eisenhauer Dec 10 '14 at 10:40

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