Suppose $T\in L(V)$. Prove that $\dim(\operatorname{null}(T))$ is equal to the number of zero singular values of T.
Proof. Suppose $T\in L(V)$. By Singular-Value Decomposition, $T$ has singular values $s_1,…,s_2$. And orthonormal bases $(e_1,…,e_n)$ and $(f_1,…,f_n)$ of V such that $$Tv=s_1 \langle v, e_1 \rangle f_1 + … + s_n \langle v, e_n \rangle f_n $$ for every $v\in V$. For each $j$, we have $ Tv_j = s_j f_j$. Hence, each $f_j$ corresponding to $s_j$ are in range $T$. Hence, nonzero $s_j$ span the range T. Hence, $\dim (\operatorname{range} (T))$ equals the number of nonzero singular values of $T$. Note that $T$ has $\dim V$ singular values.
By rank-nullity theorem, we have $\dim (V) = \dim (\operatorname{null} T) + \dim (\operatorname{range} T) $.
We have $ \dim V = n $ and $ \operatorname{range} T = n$ since number of nonzero singular values is equal to $\dim \operatorname{range} T$ . Hence $ \dim (\operatorname{null} T) = \dim V – \dim (\operatorname{range} T) = n – n = 0$. Also we have 0 number of singular values. Hence, $ \dim \operatorname{null} T = 0 = $ number of zero singular values.
Is my proof right? Thanks in advance!
Best Answer
You're correct, but needlessly elaborate. Here's an alternative:
In terms of matrices, we can write $$ A = U\Sigma V^T $$ Where $\Sigma$ is the diagonal matrix of singular values, and $U,V$ are orthogonal. Clearly, the rank of $\Sigma$ is the number of nonzero singular values.
Multiplying a matrix by an invertible matrix maintains its rank. So, $A=U \Sigma V^T$ has the same rank as $\Sigma$.