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if $A$ and $B$ are two $n\times n$ matrices with same eigenvalues such that each eigenvalue has same algebraic and geometric multiplicity. Does $A$ and $B$ are similar?

If $A$ is diagnalizable then the claim is true. But does it true even when sum of geometric multiplicity is not $n$.

Please give me a hint to start with. If given a counter example, please help me to show they are not similar. Thanks

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  • $\begingroup$ Could someone help to clarify which answer is correct plz. Answers contradict each other $\endgroup$ – Cloud JR Jun 19 at 12:59
  • $\begingroup$ Do you mean "there exists an eigenvalue having same algebraic and geometric multiplicity", or "for each eigenvalue, the algebraic and geometric multiplicity are the same"? Your wording makes it sound like you are asking the former, but this is so trivially false that one might assume that you are asking the latter. $\endgroup$ – Acccumulation Jun 19 at 15:11
  • $\begingroup$ @javadba , this is not my homework problem, i know similar matrices have same eigenvalues , with same multiplicity for each eigenvalue, and i think about the converse, but I can't prove it, so i posted it here $\endgroup$ – Cloud JR Jun 20 at 22:23
  • $\begingroup$ @Acccumulation, i am asking the latter, and i will edit it to make it precise. I'm sorry for late reply, btw $\endgroup$ – Cloud JR Jun 20 at 22:24
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    $\begingroup$ @CloudJR apologies for the confusion due to my initially incorrect answer, I hope that your question has been answered. $\endgroup$ – pre-kidney Jun 21 at 3:31
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No, it's famously false. The usual counterexample is $A=\begin{pmatrix}0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}, B=\begin{pmatrix}0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}$ for both of which $0$ has algebraic multiplicity $4$ and geometric multiplicity $2$.

The result that holds is that two matrices $A, B\in \overline{\Bbb F}^{n\times n}$, where $\overline{\Bbb F}$ is an algebraically closed field, are similar if and only if, for all $\lambda \in \overline{\Bbb F}$ and for all $m\in\Bbb N$, $\dim\ker (A-\lambda id)^m=\dim\ker(B-\lambda id)^m$. This condition trivializes to yours as soon as $n\le 3$.

If $A,B\in \Bbb F^{n\times n}$ and $\Bbb F$ isn't algebraically closed, then the same result holds, in the sense that they are similar if and only if they are similar as matrices in $\overline{\Bbb F}^{n\times n}$ or, equivalently, if and only if $\dim\ker p(A)^m=\dim\ker p(B)^m$ for all $m\in\Bbb N$ and for all irreducible polynomials $p\in\Bbb F[x]$.

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    $\begingroup$ I think a similar matrix has same eigenvalues, eigenvectors , algebraic and geometric multiplicities but the converse is not true. Your example is for the converse part. Am I right ? $\endgroup$ – nmasanta Jun 19 at 6:29
  • $\begingroup$ Your suggestion is wrong. $\endgroup$ – Saucy O'Path Jun 19 at 6:31
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    $\begingroup$ Look at the link "math.stackexchange.com/questions/8339/…" , specially in the accepted answer. $\endgroup$ – nmasanta Jun 19 at 6:33
  • $\begingroup$ @nmasanta $\small{\begin{bmatrix}1&0\\0&0\end{bmatrix}}$ and $\frac12\small{\begin{bmatrix}1&1\\1&1\end{bmatrix}}$ are similar but do not have the same eigenvectors. $\endgroup$ – amd Jun 19 at 7:03
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    $\begingroup$ @SaucyO'Path "Your suggestion is wrong". While this is generally a good answer that comment is not helpful : pls specify why. $\endgroup$ – javadba Jun 19 at 14:45
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[Begin Edit]

My initial answer was incorrect, but I believe it is interesting to explain how and why it is incorrect and provide some comments elaborating upon the correct answers posted here (I have left my initial incorrect answer unedited below).

The Jordan normal form classifies all matrices up to similarity transformations. It shows that matrices have a two step decomposition. The first step consists of the eigenvalues themselves, and the second step consists of the Jordan blocks corresponding to a given eigenvalue.

For the question under consideration here (how much information is revealed by knowing the algebraic and geometric multiplicities), distinct eigenvalues may be treated separately from one another and thus the first step in the decomposition is not essential to the question. Thus, one may focus on a single eigenvalue, and furthermore shift the eigenvalue to $0$. This leads to considering nilpotent matrices. A matrix is nilpotent if its characteristic polynomial is $x^n$, which in particular implies that it is an $n\times n$ matrix.

Any such matrix has eigenvalue $0$ with algebraic multiplicity $n$. Moreover, the eigenspace of $0$ coincides with the kernel of the matrix, from which one can see that the eigenspace is equal to the direct sum of the kernels of the Jordan blocks. In particular, the geometric multiplicity (=dimension of eigenspace) is equal to the number of Jordan blocks, since each has a kernel of dimension $1$.

Thus we see there are two cases where knowing the algebraic and geometric multiplicity is sufficient to reconstruct the matrix up to similarity: either when the algebraic and geometric multiplicities coincide (equivalent to diagonalizability), or when the geometric multiplicity is $1$.

[End Edit]


Yes, the knowledge of all the algebraic and geometric multiplicities of all eigenvalues of a matrix is sufficient to determine the matrix up to similarity transformations. This follows (and is equivalent to the existence of) the Jordan normal form.

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    $\begingroup$ See the other two answers. $\endgroup$ – amd Jun 19 at 6:57
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    $\begingroup$ Dear sir, your answer contradict , other two answer , i'm confused $\endgroup$ – Cloud JR Jun 19 at 12:55
  • $\begingroup$ You can reconstruct when the geometric multiplicity is 1 only if the matrix has only one eigenvalue $\endgroup$ – Miguel Boto Jun 20 at 4:36
  • $\begingroup$ @MiguelBoto knowing that the geometric multiplicity of each eigenvalue is $1$ implies that each eigenvalue consists of a single Jordan block with size given by the algebraic multiplicity. This is sufficient to determine the Jordan normal form, and hence recover the matrix up to similarity. $\endgroup$ – pre-kidney Jun 20 at 4:40
  • $\begingroup$ @MiguelBoto no, it is not. For example, consider the matrix $a_{i,i+1}=1$ and everywhere else $0$ (i.e., the $1$'s are right above the diagonal). This is not diagonalizable since the algebraic multiplicity of $0$ is $n$ but the geometric multiplicity is $1$. $\endgroup$ – pre-kidney Jun 20 at 5:18
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counter example: $$\begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$

$$\begin{bmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{bmatrix}$$

these two matrices have the same eigen values and same geometric multiplicity and are not similar. The geometric multiplicity of the eigen value only tells you the number of blocks in the Jordan Normal form, the size of the largest block for each eigenvalue is the first exponent $k$ such that $\dim[N(A-\lambda I)^k]=m$ where $m$ is the algebraic multiplicity of the corresponding eigenvalue $\lambda$

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    $\begingroup$ Can you prove those matrices are non similar $\endgroup$ – Cloud JR Jun 19 at 12:55
  • $\begingroup$ @CloudJR I left a few comments on some possible arguments that they are not similar in linear algebra chatroom. $\endgroup$ – Martin Sleziak Jun 19 at 13:46
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    $\begingroup$ @CloudJR they are non similar because they are block diagonal matrixes where each block is a jordan block, therefore their jordan normal form is different and there is a theorem that states that two matrices are similar if and only if their jordan normal form coincides $\endgroup$ – Miguel Boto Jun 19 at 15:00

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