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Let $A$ be an $n\times n$ symmetric positive semidefinite matrix. Let $X \in \mathbb{R}^n$

Show that if $\langle X, AX\rangle = 0,$ then $AX = 0$.

This seems like it's really simple, but I must be missing a trick.

We have $\operatorname{tr}(X^TAX) = 0$. The trace is just the sum of the eigenvalues, so could I say something about the eigenvalues all having to be zero? Would that help with the proof?

user70864
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  • Please clarify. It appears that $X, AX$ are column vectors, however your inner product appears to be for matrices. If $x,y$ are column vectors, then $x^ty$ is a number; there's no trace to be taken. – vadim123 Apr 28 '13 at 00:52
  • Your question is unclear. Is $X$ a matrix or a vector? – Julien Apr 28 '13 at 00:53
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    In any case, you might want to use that $A$ has a square root, i.e. $B$ symmetric positive semidefinite such that $A=B^2$. You don't have to go this way, but it makes things very easy. Then $(X,AX)=(BX,BX)=0$ if and only if $BX=0$, using separation if the inner product. – Julien Apr 28 '13 at 00:55
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    sorry, yes I copied the wrong inner product from the book. It doesn't specify which one to use. $X$ is vector in R^n – user70864 Apr 28 '13 at 00:57
  • So $(X,Y)=X^tY$, I guess, the Euclidean inner product. Then the square root thing works. – Julien Apr 28 '13 at 01:03

3 Answers3

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Since $A$ is symmetric, ${\bf R}^n$ has a basis consisting of eigenvectors of $A$. Express $X$ as a linear combination of these eigenvectors, then calculate $X^tAX$, then use the hypothesis about $A$ being positive semi-definite.

Gerry Myerson
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Hint: Because it is positive semi-definite, you have $$ \left<x,Ax\right> = x^TAx\geq 0 $$ Considering the properties of a symmetric positive semi-definite matrix, what does this tell you about $$ x^TA^TAx = x^TA^2x $$

Glen O
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$$rank(X^TAX) = rank(X)$$ Proof: Null space of X and $X^TAX$ is exactly the same, and hence the rank. $$Xy = 0 \implies X^TAXy = 0$$ and $$X^TAXy = 0 \implies (Xy)^TAXy = 0 \implies Xy = 0$$ Rank is also same

Therefore, this is a property for positive definite matrix A for arbitrary matrix X In our case $$X^TAX=0 \implies rank(X^TAX) = 0 = rank(X)$$

Therefore, X is a zero matrix and therefore AX=0

Abbas
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