Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is a problem from Daniel Norman's Introduction to Linear Algebra. Consider a $m\times n$ matrix, $M$, with rank $r$. We wish to show that the matrix $M^\mathrm{T}M$ also has rank $r$. Now, I can show this fact by proving that the nullspace of $M^\mathrm{T}M$ is the same as the nullspace of $M$ whereby the rank-nullity theorem will take care of the rest, but the hint of the problem proposes something confusing to me.

The hint suggests looking at the matrix $N$ obtained by exchanging two columns of $M$. It then asks "What is the relationship between $N^\mathrm{T}N$ and $M^\mathrm{T}M$?"

I can see that the two matrices are orthogonally similar, more specifically I can see that

$$E^\mathrm{T}M^\mathrm{T}ME = N^\mathrm{T}N$$ where $E = E^\mathrm{T}$ is an elementary matrix exchanging the columns in question. But I fail to see anything which leads to a resolution of this problem. Can anyone offer some help?

share|improve this question
    
@MartinArgerami Could you help me with his question please math.stackexchange.com/questions/669018/rank-of-quadratic-form ? –  juaninf Feb 14 at 17:34

1 Answer 1

up vote 1 down vote accepted

I think that the point is that you can do more than one elementary operation. Rank $r$ means that there are exactly $r$ linearly independent columns. So you can get a permutation $E$ (now a product of elementary matrices) such that $N=EME$ has its first $r$ columns linearly independent, and the last $n-r$ are linear combinations of the first $r$.

We thus have $$ N=[N_1\cdots N_n], $$ where $N_1,\ldots,N_n$ are the columns of $M$ ordered in such a way that the first $r$ are linearly independent and the last $n-r$ are linear combinations of the first $r$. So $$ N=[N_1\cdots N_r\ \sum_{k=1}^r\alpha_{1,k}N_k\cdots\sum_{k=1}^r\alpha_{n-r,k}N_k], $$ and $$ N^TN=\begin{bmatrix} N_1^TN_1&\cdots&N_1^TN_r&\sum_{k=1}^r\alpha_{1,k}N_1^TN_k&\cdots&\sum_{k=1}^r\alpha_{n-r,k}N_1^TN_k\\ \vdots& &\vdots&\vdots& &\vdots\\ N_r^TN_1&\cdots&N_r^TN_r&\sum_{k=1}^r\alpha_{1,k}N_r^TN_k&\cdots&\sum_{k=1}^r\alpha_{n-r,k}N_r^TN_k \end{bmatrix} $$

We see that the last $n-r$ columns are linear combinations of the first $r$, so the rank of $N^TN$ is at most $r$.

But the first $r$ columns are linearly independent: if $$ 0=\sum_{k=1}^r\beta_k\begin{bmatrix}N_1^TN_k\\ \vdots \\N_r^TN_k\end{bmatrix} $$ for some coefficients $\beta_1.\ldots,\beta_r$, then $$ 0=\sum_{k=1}^r\beta_kN_j^TN_k,\ \ j=1,\ldots,r. $$ But then $$ \left(\sum_{k=1}^r\beta_kN_k\right)^T\sum_{k=1}^r\beta_kN_k =\sum_{j=1}^r\beta_j\left(\sum_{k=1}^r\beta_kN_j^TN_k\right)=0 $$ So $\sum_{k=1}^r\beta_kN_k=0$, and the linearly independence of $N_1,\ldots,N_r$ implies that $\beta_1=\beta_2=\cdots=\beta_r=0$.

Thus the rank of $N^TN$ is at least $r$. But we knew it was also at most $r$, so the rank of $N^TN$ is $r$. As the rank is preserved by similarity, the rank of $M^TM$ equals that of $N^TN$ and is $r$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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