Relationship: Rank of a matrix $\leftrightarrow$ # of eigenvalues

Question

Can someone tell, why the number of the nonzero eigenvalues (counted according to their algebraic multiplicities) of a matrix of type $A^{*}A$, where $A$ is an arbitrary real or complexvalued matrix, is equal to the rank of $A$ ? Here, in step 2, it seems to me that exactly this assertion is made, and I can't quite understand it.

I get, that if $$ \text{rank}\left(A^{*}A\right)=r $$

then $$ \ker\left(A^{*}A\right)=n-r, $$ if it happens that $A^{*}A$ is an $n\times n$ matrix. But how can I conclude from this, that the multiplicity of the $0$ eigenvalue is $n-r$, i.e. that there are $n-r$ that are mapped to $0$ (couldn't it by that only linear combination of $n-r$ eigenvectors are mapped to $0$, since the kernel doesn't have to necessarily be spanned by the eigenvectors themselves, as far as I know) ?

Answer 1

The matrix $A^*A$ is self-adjoint, hence every eigenvalue is real and the matrix is diagonalizable. That means that the algebraic and geometric multiplicities of every eigenvalue agree. In particular, the nullity of $A^*A$ equals the algebraic multiplicity of the eigenvalue $0$; since the sum of the algebraic multiplicities of all eigenvalues must equal $n$ (since the characteristic polynomial of $A^*A$ must split, as all eigenvalues are real) it follows that the rank of $A^*A$ equals the sum of the algebraic multiplicities of the nonzero eigenvalues.

Now, all that remains is to show that the rank of $A^*A$ equals the rank of $A$. Since the nullspace of $A$ is contained in the nullspace of $A^*A$, we have that $\mathrm{nullity}(A)\leq\mathrm{nullity}(A^*A)$. On the other hand, if $\mathbf{v}\notin \mathrm{nullspace}(A)$, then $$0\lt \langle A\mathbf{v},A\mathbf{v}\rangle = \langle A^*A\mathbf{v},\mathbf{v}\rangle$$ so $A^*A\mathbf{v}\neq\mathbf{0}$. Thus, $\mathrm{nullspace}(A^*A)=\mathrm{nullspace}(A)$, which gives $\mathrm{nullity}(A^*A)=\mathrm{nullity}(A)$.

Since $A$ and $A^*A$ have the same number of columns, it follows as well that $\mathrm{rank}(A^*A) = \mathrm{rank}(A)$.

Answer 2

I have a hint, which I hope is not beyond your current coursework.

Did you notice $A^*A$ is a Hermitian matrix? It would be diagonalizable then...

Answer 3

Consider the singular value decomposition $A=U \Sigma V^*$. If you multiply it out in this form you immediately get an eigenvalue decomposition, $A^*A=V \Sigma^2 V^*$.

So, the nonzero squared singular values of $A$ are exactly the eigenvalues of $A^*A$, with the eigenvectors being the corresponding columns of $V$. The number of nonzero singular values of $A$ is of course the rank of $A$.

Edit: If you are not familiar with the singular value decomposition, the intuition behind it is as follows. Any matrix $A$ maps the unit sphere to an ellipsoid. In the singular value decomposition, the columns of $V$ are the orthonormal vectors on the sphere that get mapped to the axes of the ellipsoid, the columns of $U$ are the orthonormal axes of the ellipsoid, and the singular values (entries in the diagonal matrix $\Sigma$) are the scaling lengths of the ellipsoid's axes. If $A$ is rank-deficient, that means the ellipsoid is completely flattened in some directions, which is to say some of the singular values are zero.

Answer 4

I would like to give an alternate approach (though it might be the same thing).

Consider $$A \in \mathbb{C}^{(m * n)}$$ Then, consider the Single Value Decomposition (SVD) of A (since all matrices can be expressed in SVD form, we can do this in a general manner), $$ A = U \tilde{\Sigma} V^* $$ where $U$ and $V$ are unitary matrices, which also implies they are invertible, which implies they are full rank matrices.

We also know that for any matrices $C$ and $D$, $$ Rank(CD) \leq Rank(C) $$

Hence, we can say that $$ Rank(A) = Rank( U \tilde{\Sigma} V^* ) $$ $$ = Rank(\tilde{\Sigma})$$ We can write this since the ranks of U and V* are anyway maximum, so the rank of A will depend only on the rank of $\tilde{\Sigma}$.

Further, since $\tilde{\Sigma}$ is a diagonal matrix having the singular values of A (positive and zero singular values) on the principal diagonal, the rank of the matrix will be the number of positive (or non-zero, since singular values are not negative) singular values of A.

Also, the singular values of $A$ are the non-negative square roots of the eigenvalues of $A^*A$ or $AA^*$. Now, consider, $$ A^*A = (U\tilde{\Sigma} V^*)^*(U\tilde{\Sigma} V^*)$$ which reduces to, $$ A^*A = V\tilde{\Sigma}^2 V^* $$ So, $A^*A$ has the same number non-zero singular values as $A$. And $A^*A$ is Hermitian (normal) matrix, so the eigen values are the non-negative square roots of the singular values. So, by above arguments, we can conclude that the $Rank(A)$ is equal to the number of non-negative eigenvalues of A.