If $A$ is an $m\times n$ matrix, $B$ is an $n\times m$ matrix and $n

Question

The question was given in the early chapters of Linear Algebra by Hoffman & Kunze, so I am trying to give a proof with only the tools given to me so far - which are mainly row reduction and knowledge of matrix multiplication, row reduced echelon forms, row equivalence and linear independence.

I attempted a proof as per the following:

Consider $A$ as a collection (not sure if this would be the ideal expression) of $1 \times n$ row vectors, and $B$ as a collection of $n \times 1$ column vectors. Then we have that:

$$ A=\begin{bmatrix} r_1 \\ \vdots \\ r_m \end{bmatrix},\ B=\begin{bmatrix} c_1 & \cdots & c_m \end{bmatrix}. $$ Thus it follows that: $$ AB =\begin{bmatrix} r_1\cdot c_1 & \cdots & r_1\cdot c_m \\ \vdots & ~ & \vdots \\ r_m\cdot c_1 & \cdots & r_m \cdot c_m \end{bmatrix} $$ Clearly, by inspection, the rows are linearly dependent.

Since the rows of $AB$ are linearly dependent, it naturally follows that the reduced row echelon form of $AB$ contains zero rows. Hence, $AB$ is not invertible.

Would this be a mathematically sufficient proof?

Answer 1

Here's a proof that relies on matrix multiplication.

We can adjoin $m-n$ columns of zeros to $A$ and $m-n$ rows of zeros to $B$ to form $m\times m$ matrices $A', B'$. This won't affect the product $AB$, meaning $A'B'=AB$.

Then we'll have something like

$$ \left[\begin{array}{c|c} A & 0 \end{array} \right] \left[\begin{array}{c} B\\ \hline 0 \end{array}\right]=A'B'$$

Since $A'$ has a column of zeros, $\det A'=0$, so $\det{A'B'}=0$ .

But since $AB=A'B'$, we have $\det{AB}=\det{A'B'}=0$, which means that $AB$ is not invertible.

Answer 2

As has been mentioned in the comments, your approach does not make a for complete and proper proof. You may proceed as follows:

We have from here, for example, that $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$. This you could try to prove using the tools you already know.

In your problem we have that $\operatorname{rank}(A)\leq n$ and $\operatorname{rank}(B)\leq n$ implying that $\operatorname{rank}(AB) \leq n$, but $AB$ is $m$ by $m$ with $n<m$ hence $AB$ can not have full rank which in turn means that it is not invertible.

Answer 3

I have a short proof for this .

Assume AB = C and C is invertible .

Now we know that to convert any matrix into its reduced row echelon form , we multiply it through a series of elementary matrices . Let the cumulative product of these elem. matrices be P (mxm matrix)

PA = R (rref(A))

=> PAB = RB = PC

now consider R , there are more number of rows than columns , at most n pivots can be there which is less than m => there has to be a row with no pivots => a zero row

As a result RB (a matrix mxm) has atleast one zero row . This implies that RB is not invertible , but P is invertible (product of elem matrices) . Therefore our assumption that C is invertible is wrong and so a contradiction occurs.

P.S. : I am new to writing proofs so don't have a lot of exposure of putting sentences into symbols !

Answer 4

Here's a proof involving linear transformations. Let $F$ be a field.

Given that, $A$ is an $m\times n$ matrix. This means that A is the matrix of some linear transformation $U: F^n \to F^m$ with respect to some bases of $F^n$ and $F^m$.

Again, $B$ is an $n\times m$ matrix means that B is the matrix of some linear transformation $T: F^m \to F^n$ with respect to some bases of $F^m$ and $F^n$.

Now, consider the linear transformation $UT: F^m \to F^m$. We prove that $UT$ is not invertible, which will in turn imply that the matrix of $UT$ (with respect to some basis of $F^m$), namely $AB$, is not invertible.

The important thing to notice is the given information $\textbf{n < m}$.

Consider a basis $\mathcal{B}$ for $F^m$, given by $$\mathcal{B} = \{\alpha_1, \alpha_2, \ldots, \alpha_m\}$$ Next, consider the set $\{T(\alpha_1), T(\alpha_2), \ldots, T(\alpha_m)\}$. Clearly, this set is linearly dependent as the range of $T$ (which is $F^n$) has dimension $n$, which is less than $m$, the number of vectors in our set.

$$\implies b_1T(\alpha_1) + b_2T(\alpha_2) + \ldots + b_mT(\alpha_m) = 0 \; where \; b_1, b_2, \dots, b_m \in F \; are \; not \; all \; 0 \ldots (1) $$

$$\implies b_1\alpha_1 + b_2\alpha_2 + \ldots + b_m\alpha_m \neq 0 \dots (2)$$ (as $\{\alpha_1, \alpha_2, \ldots, \alpha_m\}$ is a linearly independent set, being a basis).

$$$$ Now, $$UT(b_1\alpha_1 + b_2\alpha_2 + \ldots + b_m\alpha_m)$$ $$ = U(T(b_1\alpha_1 + b_2\alpha_2 + \ldots + b_m\alpha_m))$$ $$ = U(b_1T(\alpha_1)+b_2T(\alpha_2)+\ldots + b_mT(\alpha_m))$$ $$ = U(0) \;[from \; (1)] = 0 $$

Thus, $UT(nonzero\; vector) = 0$ means that $UT$ has a non trivial null-space, which in turn implies that $UT$ is not injective. Thus, $UT$ is not invertible. $\blacksquare$