Let A be a real $m \times n$ matrix. Prove that there are orthogonal matrices $P$ in $O_m$, and $Q$ in $O_n$ such that $PAQ$ is diagonal, with non-negative diagonal entries.

If we prove the statement for $n \times n$ invertible matrices we can easily prove the statement for arbitrary $m \times n $ matrices.

The $n \times n $ case can be restated as:

Pseudo-orthogonal Lemma: Given a invertible linear transformation $T$ from $\textbf{R}^a$ to $\textbf{R}^a$ there exists $u\ne \textbf{0} \in \textbf{R}^a$ such that: $\langle u,w\rangle=c\langle T(u),T(w)\rangle$ for all $w \in \textbf{R}^a$ where $c = \frac{\langle u,u\rangle}{\langle T(u),T(u)\rangle}$

OR stated differently: $ \exists u\ne \textbf{0}$ such that $T(W^{\bot})\bot\ T(W)$ where $W=span\{u\}$

We show how the problem can be solved assuming the Pseudo-orthogonal Lemma.

Sub-Lemma: Given an invertible $n\times n $ matrix $B$ there exists $P$ in $O_n$ and $Q$ in $O_n$ such that $P A Q$ is diagonal, with positive entries.

Proof: We prove this by induction on the order of the matrix.

Base case: $n=1$, the sub-lemma is trivially true.

Induction hypothesis: Given an invertible ${k-1}\times{k-1} $ matrix $B$ there exists $P^{(k-1)}$ in $O_{k-1}$ and $Q^{(k-1)}$ in $O_{k-1}$ such that $P A Q$ is diagonal, with positive diagonal entries.

Inductive step: By Pseudo-orthogonal Lemma $\exists w\ne \textbf{0}$ such that $T_B(W^{\bot})\bot\ T_B(W)$ where $W=span\{w\}$. Let $\{w,v_1,..v_{k-1}\}$ be an orthonormal basis of $ \textbf{R}^n$ and let $\{\frac{T_B(w)}{||T_B(w)|| },z_1,..z_{k-1}\}$ be an orthonormal basis of $ \textbf{R}^n$ where $T_B$ is the linear transformation whose matrix with respect to standard basis is $B$.

Now let $T_B^{(k-1)}$ be the linear transformation from $W^{\bot}$ to $Im(T_B^{(k-1)})$ where $T_B^{(k-1)}(x)=T_B(x)$. Since $T_B(W^{\bot})\bot\ T_B(W)$, $\langle T_B(x), T_B(w)\rangle=0\ \forall\ x \in W^\bot$,

hence $\langle T_B^{(k-1)}(x), T_B(w)\rangle=0\ \forall\ x \in W^\bot$ thus $Im(T_B^{(k-1)})=T_B(W)^\bot$.

As $B=\{z_1,..z_{k-1}\}$ is an orthonormal basis of $T_B(W)^\bot$, we can write $T(v_i)=\sum_kd_{ik}z_k$.

Thus $ \begin{bmatrix} -&-&T_B(v1)&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&T_B(v_{k-1})&-&- \\ \end{bmatrix} = \begin{bmatrix} d_{11}&d_{12}&-&-&- \\ d_{21}&d_{22}&-&-&- \\ -&-&-&-&- \\ -&-&-&-&- \\ \end{bmatrix} \begin{bmatrix} -&-&z_1&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&z_{k-1}&-&- \\ \end{bmatrix} $

Let $D=\begin{bmatrix} d_{11}&d_{12}&-&-&- \\ d_{21}&d_{22}&-&-&- \\ -&-&-&-&- \\ -&-&-&-&- \\ \end{bmatrix}$

Now by applying induction hyphothesis on $D$ we have $Q_1,Q_2$, real $k-1\times k-1$ orthogonal matrices such that $\Lambda^{(k-1)}=Q_1DQ_2$ is diagonal with non negative entries.

$B^t= \begin{bmatrix} -&-&T_B(e_1)&-&- \\ -&-&T_B(e_2)&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&T_B(e_k)&-&- \\ \end{bmatrix} = \begin{bmatrix} -&-&T_B(w)&-&- \\ -&-&T_B(v1)&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&T_B(v_{k-1})&-&- \\ \end{bmatrix} \begin{bmatrix} -&-&w&-&- \\ -&-&v_1&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&v_{k-1}&-&- \\ \end{bmatrix}^{-1} $

We note $Q_3=\begin{bmatrix} -&-&w&-&- \\ -&-&v_1&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&v_{k-1}&-&- \\ \end{bmatrix}^{-1}$ is a $k \times k$ real orthogonal matrix as $\{w,v_1,..$$v_{k-1}$$\}$ is an orthonormal basis.

$ \begin{bmatrix} -&-&T_B(w)&-&- \\ -&-&T_B(v_1)&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&T_B(v_{k-1})&-&- \\ \end{bmatrix} =M= \begin{bmatrix} ||T_B(w)||&0&..&..&0\\ 0&d_{11}&d_{12}&-&-\\ 0&d_{21}&d_{22}&-&- \\ 0&-&-&-&- \\ 0&-&-&-&- \\ \end{bmatrix} \begin{bmatrix} -&-&\frac{T_B(w)}{||T_B(w)|| }&-&- \\ -&-&z_1&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&z_{k-1}&-&- \\ \end{bmatrix} $

We note $Q_4=\begin{bmatrix} -&-&\frac{T_B(w)}{||T_B(w)|| }&-&- \\ -&-&z_1&-&- \\ -&-&..&-&- \\ -&-&..&-&- \\ -&-&z_{k-1}&-&- \\ \end{bmatrix}$ is a $k \times k$ real orthogonal matrix as $\{\frac{T_B(w)}{||T_B(w)|| },z_1,..z_{k-1}\}$ is an

Next we note $Q_5=\begin{bmatrix} 1&\\ 0&Q_1^{-1}\\ \end{bmatrix}$ and $Q_6=\begin{bmatrix} 1&\\ 0&Q_2^{-1}\\ \end{bmatrix}$ are also orthogonal matrices and $\begin{bmatrix} ||T_B(w)||&0&..&..&0\\ 0&d_{11}&d_{12}&-&-&- \\ 0&d_{21}&d_{22}&-&-&- \\ 0&-&-&-&-&- \\ 0&-&-&-&-&- \\ \end{bmatrix} = Q_5 \begin{bmatrix} ||T_B(w)||&\\ 0&\Lambda^{(k-1)} \\ \end{bmatrix} Q_6 $

Thus $B^t=MQ_3 = Q_5 \begin{bmatrix} ||T_B(w)||&\\ 0&\Lambda^{(k-1)} \\ \end{bmatrix} Q_6Q_4Q_3$.

Since $\Lambda = \begin{bmatrix} ||T_B(w)||&\\ 0&\Lambda^{(k-1)} \\ \end{bmatrix}$ is a diagonal matrix with positive entries and since $Q_7=Q_6Q_4Q_3$ is an orthogonal matrix we have proved the Sub-Lemma.

  • $\begingroup$ This is equivalent to the existence of the singular value decomposition. Many linear algebra texts contain a proof $\endgroup$ Aug 8, 2020 at 14:25
  • $\begingroup$ Also, it's not clear what exactly your question is. Based on your proof of the lemma, it seems unlikely that you want to be given a proof of the statement. So with that said, what exactly are you asking for? $\endgroup$ Aug 8, 2020 at 14:26
  • $\begingroup$ @BenGrossmann I have solved the question assuming a certain hypothesis which I have called the pseudo-orthogonal lemma. I am requesting for a proof of the pseudo-orthogonal lemma which I have assumed without proof and do not know how to prove $\endgroup$ Aug 9, 2020 at 4:06
  • $\begingroup$ Is there a reason that you are trying to do things this way instead of reading about singular value decomposition in a textbook? $\endgroup$ Aug 9, 2020 at 9:57
  • $\begingroup$ In the future, please make the nature of your question clearer. It is difficult to look at your question as it is written and understand that this is what you want. $\endgroup$ Aug 9, 2020 at 10:04

1 Answer 1


Let $T^*$ denote the adjoint to $T$ (i.e. the transformation corresponding to the transpose of the matrix of $T$). We find that $\langle Tu,w \rangle = \langle u,T^*w \rangle$ holds for all $u,w \in \Bbb R^a$. Note that in the usual approach, the SVD is proved as a consequence of the spectral theorem and the fact that $(T^*T)^* = T^*T$ (i.e. $T^*T$ is self-adjoint). However, given your unconventional approach to this problem (i.e. your decision not to simply read a textbook), I assume that you want to avoid this.

With that in mind, I begin with the following claim.

Claim: There exists a unit-vector $u$ for which $\|Tu\| = \max_{x \in \Bbb R^a,\|x\| = 1} \|Tx\|$.

This is a consequence of the fact that the unit-ball $\{x: \|x\| = 1\}$ is compact, and the function $f(x) = \|Tx\|$ is continuous. I now claim that as a consequence, it holds that for any $w \perp u$, it holds that $Tw \perp Tu$. In other words, $u$ satisfies the condition of the pseudo-orthogonal lemma.

Indeed, suppose for the purpose of contradiction that $w$ is a unit vector with $w \perp u$, but $\langle Tu,Tw\rangle \neq 0$. It follows that \begin{align} \| T(\cos \theta u + \sin \theta w)\|^2 &= \langle T(\cos \theta u + \sin \theta w), T(\cos \theta u + \sin \theta w)\rangle\\ &= \|Tu\|^2\cos^2\theta + \|Tw\|^2\sin^2 \theta + 2 \langle Tu, Tw\rangle \sin \theta \cos \theta \\ & = \|Tw\|^2 + (\|Tu\|^2 - \|Tw\|^2)\cos^2 \theta + 2 \langle Tu, Tw\rangle \sin \theta \cos \theta \\ & = a + b\cos^2 \theta + c \sin \theta \cos \theta \\ & = a_0 + b_0 \cos(2\theta) + c_0 \sin(2 \theta), \end{align} where $c_0 \neq 0$. By the maximality of $\|Tu\|$, it should hold that the function $$ f(\theta) = a_0 + b_0 \cos(2\theta) + c_0 \sin(2 \theta) $$ attains a maximum at $\theta = 0$. However, we compute $$ f'(\theta) = -2b_0\sin(2 \theta) + 2c_0 \cos(2\theta) \implies f'(0) = 2c_0 \neq 0, $$ which means that $f$ does not attain a maximum at $\theta = 0$, which is a contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.