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The question Im encounter is given a matrix:

$$ A^TA = \begin{bmatrix} 1 & -2 & -4\\ -2 & 4 & 8\\ -4 & 8 & 42 \end{bmatrix} $$

I need to tell how many eigenvalues of given matrix is positive? zero or negative. From this matrix I can find two information

  1. the rank is 2
  2. it's symmetric matrix

The answer for this question is 2 positive e-values because rank=2 ,1 zero e-value because it's singular.

But I don't understand why rank and e-value related? How to know it's singular without compute determinate.

And I also want to know does $A^TA$ has same eigenvector as $A$?

Yiffany
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2 Answers2

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A positive semi-definite matrix is diagonalizable and its eigenvalues are non-negative, hence its rank equals its number of positive eigenvalues.

Details of the "hence": since Similar Matrices have the same rank, if $B=PDP^{-1}$ with $D=\operatorname{diag}(\lambda_1,\dots,\lambda_r,0,\dots,0)$ and $\lambda_1,\dots,\lambda_r\ne0$, then $\operatorname{rank}(B)=\operatorname{rank}(D)=r$.

Anne Bauval
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    And your $A$ may have no eigenvector, but I think that additional question should have been posted separatly. – Anne Bauval Oct 03 '22 at 06:26
  • How do you know it's positive semi-definite without knowing all e-values? The defination is all e-value >= 0 , it's positive semi-def. – Yiffany Oct 03 '22 at 06:54
  • https://math.stackexchange.com/questions/2450437/for-any-m-times-n-matrix-a-the-matrices-ata-and-aat-are-positive – Anne Bauval Oct 03 '22 at 07:05
  • where can I find the proof of your answer? – Yiffany Oct 03 '22 at 08:03
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    Of which part? 1. the fact that $A^TA$ is positive semi-definite? (with which definition?), or 2. that a real symmetric matrix is diagonalizable? or 3. that the eigenvalues of a positive semi-definite matrix (with which definition) are non-negative? or 4. that this entails that the rank is number of positive eigenvalues? or 5. something else? – Anne Bauval Oct 03 '22 at 08:42
  • #4 , why rank=num of positive e-value? thank you!!! – Yiffany Oct 03 '22 at 09:07
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Sylvester's Law of Inertia says two positive eigenvalues and a single zero eigenvalue.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 4 & 0 & 1 \\ 2 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 4 & 8 \\ - 4 & 8 & 42 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 4 & 2 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 26 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 0 & 1 \\ - 4 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 26 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 4 & 8 \\ - 4 & 8 & 42 \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Will Jagy
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