Properties of matrices, including span , vectors and products

Question

I am studying matrices and their properties, and i want to make sure that my understanding of them is as complete as it can be.

Please correct me if i'm saying anything wrong .

If the product of two matrices AB is a square matrix, then both A and B are square matrices.

Moreover, if we have 4 vectors in R3, then these four vectors must span all the space.

This leads to the third property , where the number of row columns of a matrix gives the dimension of the span of the column vectors.

Is my understanding of these three properties correct? Please tell me if im wrong ! Thanks!

Answer 1

You are wrong in each case.

1. $\displaystyle\begin{pmatrix}a&b\end{pmatrix}.\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}ac+bd\end{pmatrix}$, which is a square matrix.
2. The vectors $(1,0,0)$, $(2,0,0)$, $(3,0,0)$, and $(4,0,0)$ do not span $\mathbb{R}^3$.
3. My guess is that when you wrote “the number of row columns of a matrix”, you meant “the number of columns of a matrix”. Think about the previous answer.

Answer 2

You need to go back to basics. Both your assertions are false more often than they are true.

The product of a $1 \times n$ and an $n \times 1$ matrix is a $1 \times 1$ matrix.

The vectors $(0,0,0)$, $(1, 0, 0)$, $(2,0,0)$, $(3, 0, 0)$ do not span $\mathbb{R}^3$.

You should now see why your third assertion is false.

Answer 3

In general an $m\times p$ matrix multiplied by an $p\times n$ matrix gives an $m\times n$ matrix. For matrix multiplication to be possible, we just need the total number of columns of the first matrix to equal the total number of rows of the second matrix (why?)

So letting $m=n$ in the above has the product of two nonsquare matrices give an $n\times n$ square matrix. Also consider an $m\times n$ matrix, $A$, and its transpose, $A^t$, which will be a $n\times m$ matrix (again why?). These multiply as $A\cdot A^t$ giving an $m\times m$ square matrix, and as $A^t\cdot A$ giving an $n\times n$ square matrix. As an example, let $A$ be the $3\times 2$ matrix with its $2\times 3$ transpose $A^t$: $$ A= \begin{bmatrix} 2 & 3 \\ 5 & 7 \\ 11 & 13 \\ \end{bmatrix}\quad A^t= \begin{bmatrix} 2 & 5 & 11 \\ 3 & 17 & 13 \\ \end{bmatrix} $$

$$A\cdot A^t= \begin{bmatrix} 13 & 31 & 61 \\ 31 & 74 & 146 \\ 61 & 146 & 290 \\ \end{bmatrix} \quad A^t\cdot A= \begin{bmatrix} 150 & 184 \\ 184 & 227 \\ \end{bmatrix} $$ Here $A\cdot A^t$ giving an $3\times 3$ square matrix, and as $A^t\cdot A$ giving an $2\times 2$ square matrix.

Three nonzero vectors, $v_1$, $v_2$, $v_3\in\mathbb{R}^3$, span the whole of $\mathbb{R}^3$ only if they are linearly independent (here $3$ is the dimension of the space, and in general you need $n$ linearly independent vectors to span an $n$-dimensional space). That is if the only solution to $$\lambda_1v_1+\lambda_2v_2+\lambda_3v_3=0,\quad \lambda_i\in\mathbb{R}$$ is the trivial one having all scalars $\lambda_1=\lambda_2=\lambda_3=0$. Otherwise we could rearrange the equation to write one of the vectors $v_i$ in terms of the other two. If we had two linearly independent vectors in $\mathbb{R}^3$, then they would span a $2$-dimensional subspace of $\mathbb{R}^3$. We call the three linearly independent vectors a maximal spanning set for this reason, and if this were the case adding another vector $v_4$ to the set $S=\{v_1,v_2,v_3\}$ would have $v_4$ a linear combination of these vectors, by virtue of the fact that they already span $\mathbb{R}^3$. This wouldn't necessarily be the case if any of the vectors in $S$ were linearly dependent, as then they wouldn't span $\mathbb{R}^3$, so adding $v_4$ either makes $S$ into a spanning set or it doesn't (why-consider all the different scenarios?)