Ytukyg

Let $A$ be a $3times 2$ matrix and $B$ be a $2times 3$ matrix. Show that $C = Acdot B$ is a singluar matrix.

Not sure how to proceed with this. How can it be always be singular if we are randomly multiplying two matrix

The beauty of linear algebra is that it gives us the language to explain the same situation in many different terms.

For example, an $p$ by $p$ matrix $D$ being singular is equivalent to

• $det(D)=0$


• $rank(D)<p$


• Their existing some vector other than the zero vector in $null(D)$
  

Using the second equivalence $rank(D)<p$ this tells us that to show a matrix $C=AB$ is singular, we must show that $rank(C)<3$ (since the product of $Ain R^{3x2}$ and $Bin R^{2x3}$ is in $R^{3x3}$).

Rectangular, non square matrices have a maximum rank of their smallest dimension (I mean spatial dimension rather than vector space dimension).

i.e. $$Ain R^{a x b},quad rank(A)le a quad (ale b)$$

We can use the fact that $rank(AB)le min(rank(A), rank(B))$.

We have

$$rank(A)le 2 \ rank(B)le 2$$

So

$$rank(C)=rank(AB)le 2 < 3$$

So $C$ is singular.