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Let $A$ be a square real symmetric matrix with rank $1$ , then can all the diagonal entries of $A$ be $0$ ? I know that real symmetric matrices are diagonalizable . Also if all the diagonal entries be $0$ then sum of all the eigenvalues will be $0$ . But so what ? Please help . Thanks in advance

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    $\begingroup$ What does "rank $1$" tell you about the matrix? $\endgroup$ – Daniel Fischer Feb 6 '16 at 11:41
  • $\begingroup$ If an $n \times n$ Matrix has rank 1, what is the dimension of its kernel? What are the dimensions of the eigenspace for the eigenvalue $0$? Are there other eigenvalues? $\endgroup$ – Roland Feb 6 '16 at 11:42
  • $\begingroup$ @Roland : yes , but what does the dimension of the kernel tell me ? $\endgroup$ – user228169 Feb 6 '16 at 11:57
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    $\begingroup$ No, rank $1$ means that the dimension of the range is $1$, and hence the kernel has dimension $n-1$. Not every (nonzero) vector is an eigenvector if $n > 1$. So there are $n-1$ eigenvalues $0$, and one nonzero eigenvalue. $\endgroup$ – Daniel Fischer Feb 6 '16 at 12:04
  • $\begingroup$ @DanielFischer : ah yes , yes right . $\endgroup$ – user228169 Feb 6 '16 at 12:06
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The quick answer to your question: note that the only diagonalizable matrix whose eigenvalues are all $0$ is the zero-matrix, and that a rank $1$ matrix can have at most one non-zero eigenvalue.


Another approach:

Note that any rank $1$ matrix can be written in the form $uv^T$ for column vectors $u,v$, and that a rank-$1$ matrix will be symmetric if and only if it can be written in the form $uu^T$ for some vector $u$.

Now, if $A = uu^T$, then the diagonal entries are given by $A_{ii} = u_i^2$.

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The matrix is diagonalizable for the spectral's theorem. Indeed the dimension of the eigenspace for the eigenvalue of $0$ is $n-1$. Note that the eigenspace for the eigenvalue of $0$ is the $\ker$ of function.

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  • $\begingroup$ This looks like a hint, rather than an answer. (Namely it's missing that the trace is the sum of the diagonal elements of $A$ and the sum of the eigenvalues of $A$. If $\lambda$ is the nonzero eigenvalue, the latter is $(n-1)*0 + 1*\lambda$, thus the sum of the diagonal elements is equal to $\lambda \neq 0$, regardless if $A$ is in diagonal form or not. $\endgroup$ – Roland Feb 6 '16 at 12:47
  • $\begingroup$ @Roland yes..... it's a little hint $\endgroup$ – Domenico Vuono Feb 6 '16 at 12:49

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