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I have been given two matrices that are defined as follows: $$A = \left(\begin{array}{rrrr} -1&2&-3&-1\\% 8&-7&12&4\\% 6&-6&10&3\\% 2&-2&3&2\\% \end{array}\right), B= \left(\begin{array}{rrrr} -1&-4&7&-6\\% 0&11&-25&25\\% -4&8&-25&28\\% -4&4&-16&19\\% \end{array}\right) \in \mathbb{Q}^{4 \times 4}.$$

I need two prove or disprove that the matrices A and B are similar.

The matrices A and B have the same eigenvalue, which are equal to $1$. Furthermore, I found out that the eigenspaces of the different matrices are not equal. The eigenspace for a given matrix A and eigenvalue a is defined as the solution set of: $$V_a(A) = \mathbb{L}(A - a \cdot E_n, 0)$$ The geometric multiplicity for a matrix $A$ of an eigenvalue $a$ is defined as the dimension of the eigenspace: $$g_a(A) = dim (\mathbb{L}(A - a \cdot E_n, 0))$$ For the eigenspaces and geometric multiplicity of the matrices A and B, it follows: $$V_1(A) = \mathbb{L}(A - 1 \cdot E_4, 0) = \{x_2 \cdot \left(\begin{array}{rrr} 1\\% 1\\% 0\\% 0\\% \end{array}\right) + x_3 \cdot \left(\begin{array}{rrr} -3/2\\% 0\\% 1\\% 0\\% \end{array}\right) + x_4 \cdot \left(\begin{array}{rrr} -1/2\\% 0\\% 0\\% 1\\% \end{array}\right) | x_2, x_3, x_4 \in \mathbb{Q}\}$$ $$\Rightarrow dim (\mathbb{L}(A - 1 \cdot E_4, 0)) = 3$$ $$\Rightarrow g_1(A) = 3$$ $$V_1(B) = \mathbb{L}(B - 1 \cdot E_4, 0) = \{x_3 \cdot \left(\begin{array}{rrr} -3/2\\% 5/2\\% 1\\% 0\\% \end{array}\right) + x_4 \cdot \left(\begin{array}{rrr} 2\\% -5/2\\% 0\\% 1\\% \end{array}\right) | x_3, x_4 \in \mathbb{Q}\}$$ $$\Rightarrow dim (\mathbb{L}(B - 1 \cdot E_4, 0)) = 2$$ $$\Rightarrow g_1(B) = 2$$ Because the geometric multiplicity of both matrices are not equal I would assume that matrices A and B are not similar. But how do I prove this formal. I know if two matrices (e.g. A and B) are similar, it follows: $$A = T^{-1} \cdot B \cdot T \, \text{for} \, T \in GL_n(\mathbb{Q})$$ How would I use this definition to disprove the given statement.

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If $A$ and $B$ are similar matrices, then for every eigenvalue $a$, the eigenspaces $V_a(A)$ and $V_b(B)$ are isomorphic (and hence have the same dimension). In your notation, if $A = T^{-1}BT$, then $v$ is an eigenvector of $A$ if and only if $T\cdot v$ is an eigenvector of $B$ (verify this). So, the isomorphism is given by $v \mapsto T\cdot v$, from $V_a(A)$ onto $V_a(B)$.

However, you showed that these spaces have different dimensions, hence $A$ and $B$ are not similar. (this is just the contrapositive of the statement above).


Alternatively, if you know about Jordan Canonical form, you can answer this question as follows: $\dim V_a(A) = 3$ means the JCF of $A$ has $3$ Jordan blocks corresponding to the eigenvalue of $1$. However, $\dim V_a(B) = 2$ means that in the JCF of $B$ there are only $2$ Jordan Blocks corresponding to the eigenvalue $1$.

Since similar matrices (whose characteristic polynomials split) have the same JCF (up to permutation of the blocks), it follows that $A$ and $B$ are not similar.

By the way, in your particular example, the Jordan canonical forms are: \begin{align} J_A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \quad \text{and} \quad J_B = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align} respectively (which are not similar)

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The similarity transformation preserves the geometric multiplicities of the eigenvalues, because it maps eigenvectors to eigenvectors. To see why, check that $T(v)$ is an eigenvector of $TAT^{-1}$ just when $v$ is an eigenvector of $A$.

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  • $\begingroup$ In the OP's notation, one could show that $T$ restricted to $V_a(A)$ maps it isomorphically onto $V_a(B)$. $\endgroup$ – Matthew Leingang Jun 24 at 15:30

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