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How to prove that if there exists a vector $\underline a\ne 0$ such that $Var(\sum_i a_iX_i)=0$ the covariance matrix is not invertible?

Mostafa Ayaz
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Radosław Gdula
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1 Answers1

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The condition $Var(\sum_{i=1}^n a_iX_i)=0$ can also be written in another form: $$Var\{a^TX\}=0$$which is equivalent to $${E\{(a^TX)^2\}-E^2\{a^TX\}=0 \\E\{a^TXX^Ta\}-E^2\{a^TX\}=0 \\a^TE\{XX^T\}a-E\{a^TX\}E\{X^Ta\}=0 \\a^TE\{XX^T\}a-a^TE\{X\}E\{X^T\}a=0 \\a^T[E\{XX^T\}-E\{X\}E\{X^T\}]a=0 \\a^T\Sigma_X a=0 }$$which means that the covariance matrix $\Sigma_X$ has a $0$ eigenvalue with the corresponding eigenvector $a$ and hence, is not invertible.

Mostafa Ayaz
  • 33,056
  • Could you explain why $a^T \Sigma _X a=0$ gives us the eigenvalue? I didnt recall finding eigenvalues by simply multiplying any vecor by matrix and then by its tranpose. Thanks in advance. – Radosław Gdula Apr 17 '20 at 18:45
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    This is because since $\Sigma_X$ is symmetric, you can write $\Sigma_X=AA^T$ for some matrix $A$, which leads to $A^Ta=0$ or $AA^Ta=0$. – Mostafa Ayaz Apr 17 '20 at 19:30
  • I think i'm starting to get it, but i'm still confused a bit $a^T\Sigma_X a=0 \a^TAA^T a=0 \(a^TA)(a^TA)^T=0 $

    So i get that $a^TA=0$ but what now? It means $0$ is an eigenvalue of $A$ but does it imples being eigenvalue of $\Sigma_X$ as well?

     – Radosław Gdula Apr 18 '20 at 12:02
  • Then it is implied that $A^Ta=0$ and consequently $AA^Ta=0$. – Mostafa Ayaz Apr 18 '20 at 12:40