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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!

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    $\begingroup$ You don't need the final paragraph (in fact, it's incorrect, because there's no reason the range has to be $n$ dimensional). Once you have noted that $\textrm{dim(range}\,T)$ is equal to the number of nonzero singular values, you have: $\textrm{dim(null}\,T) = \textrm{dim}(V) - \textrm{dim(range}\,T) = n - \textrm{number of nonzero singular values} = \textrm{number of zero singular values}$. $\endgroup$ – BaronVT Dec 16 '14 at 3:40
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    $\begingroup$ Thanks, so i guess my proof is fine other than that? $\endgroup$ – needhelp Dec 16 '14 at 3:43
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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 $\Sigma$.

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  • $\begingroup$ Thanks for providing me the alternative! $\endgroup$ – needhelp Dec 16 '14 at 4:04
  • $\begingroup$ Is the number of zero eigenvalues equal to the number of zero singular values ? $\endgroup$ – Fareed Abi Farraj Jul 15 '19 at 7:53
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    $\begingroup$ @FareedAF If you're referring to geometric multiplicity, then no. If you're referring to algebraic multiplicity, then yes. $\endgroup$ – Omnomnomnom Jul 16 '19 at 14:26

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