If this matrix is diagonalizable then what are the values of $a_1$ through $a_6$? $$\begin{bmatrix} 3&a_1&a_2&a_3\\ 0&3&a_4&a_5\\ 0&0&3&a_6\\ 0&0&0&3 \end{bmatrix} $$
I understand that the eigenvalues are $3$, but how do you solve for the variables?
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the eigenspace will be a 4x4, because there are 4 eigenvalues which are all 3 – jared smith Dec 13 '13 at 23:11
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When all else fails, actually diagonalize if you cannot figure out the Jordan Normal Form from the nice answer. – Amzoti Dec 13 '13 at 23:33
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@jaredsmith: you've answered your own question by stating that the eigenspace of the only eigenvalue, $3$, has dimension $4$; note that this is equal to the multiplicity of the eigenvalue. This is equivalent to diagonalizability and so we are done. – Christopher K Dec 14 '13 at 00:23
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Hint: Let $E_A(3)=\{v\in\mathbb{R}^4:Av=3v\}$ be the eigenspace relative to $3$ ($A$ is your matrix). What's the condition for diagonalizability? The dimension of $E_A(3)$ should be…
What's the algebraic multiplicity of the eigenvalue $3$? The characteristic polynomial is $(3-X)^n$ (guess what's $n$).
In this particular case, the problem is easier: the only diagonal matrix to which $A$ can be similar is $$D=\begin{bmatrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ Now, can you compute $SDS^{-1}$?
egreg
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@jaredsmith Done, but you should remember the condition for diagonalizability. – egreg Dec 13 '13 at 23:54
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the multiplicity is 4, and the condition is A=PDP^-1, but how does this help me know what the values of a1-a6? – jared smith Dec 14 '13 at 00:04
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The characteristic polynomial is $(\lambda - 3)^4$, hence the only possible eingevalue is $3$. Therefore, over the basis which makes this matrix diagonal, there are only $3$'s on the diagonal. Conclude from this that $a_1 = \cdots = a_6 = 0$.
Hope that helps,
Patrick Da Silva
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