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Why is it obvious that a matrix A has an eigenvalue $\lambda_4=0$ if we know that A is symmetric 4x4 matrix, has a double eigenvalue 5 and another simple eigenvalue of -1, and rank 3?

randomname
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  • Nope the statement above is wrong, consider this $\text{diag}(5,5,-1,a,b)$ where $a,b\in \mathbb{R}$ are arbitrary. – K.K.McDonald May 25 '20 at 05:58
  • I have modified the information. – randomname May 25 '20 at 06:00
  • Does this help? https://math.stackexchange.com/questions/1349907/what-is-the-relation-between-rank-of-a-matrix-its-eigenvalues-and-eigenvectors – Kabouter9 May 25 '20 at 09:27
  • I don't think it answers my question since in the information I have, the matrix has full rank. According to the article, if 0 is an eigenvalue then the rank is n-1. This seems conflicting to me for answering my question. – randomname May 27 '20 at 23:38

1 Answers1

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Hint the eigenspace relative to zero corresponds to the solution space or $Ax=0$. So..

InsideOut
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