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Let

$$A=\begin{bmatrix} 0&1&-1\\ -1&-2&5\\ 0&0&3 \end{bmatrix} $$

i) Prove that $A$ is similar to a diagonal matrix, but not unitarily similar to a diagonal matrix.

ii) Find two $A-$invariant subspaces $V_1$ and $V_2$ such that $\mathbb{R}^3$ is the direct sum $V_1$ and $V_2$.

I know that $spec(A)=\{-1,3\}$ and its characteristic polynomial is $p(x)=(x-3)(x+1)^2$. We also have Schur's theorem , but I don't know how to solve the probelm. I appreciate any help.

1 Answers 1

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It is clear that $\dim \ker(A + I) = 3 - \operatorname{rank}(A + I) = 1$ so the geometric multiplicity of the eigenvalue $-1$ is one and $A$ is not diagonalizable. A direct sum decomposition of $\mathbb{R}^3$ is given by $$\mathbb{R}^3 = \ker(A + I)^2 \oplus \ker(A - 3I)$$ and each (generalized) eigenspace is $A$-invariant.

The matrix $A$ is not orthogonally similar to a diagonal matrix because it is diagonalizable so it is definitely not orthogonally diagonalizable.

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    sorry I didn't get why $A$ is similar to a diagonal. Yes it's eigenvalues are -1,3, and it's jordan form would be $$\begin{bmatrix} -1&-1&0\\0&-1&0\\0&0&3 \end{bmatrix}$$. – 2017-01-24
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    @Parisina: Yeah, sorry, the rank is $1$ and not $2$. I've fixed my answer. – 2017-01-24
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    the questions says that $A$ is similar to a diagonal but not unitarily simialr to a diagonal matrix. However, since algebraic multiplicity of eigenvalue $-1$ is not equal to its geometric multiplicity, the matrix $A$ is not diagonalizable . I think there is a problem. Right? – 2017-01-24
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    Indeed, the question as stated is wrong. – 2017-01-24