How do I know the span of a matrix

Question

I have a marix that is $$R=\pmatrix{1&1&1\\1&1&1+2a\\0&2&2+2b}$$

and the span of R is span=$\left\{\pmatrix{1\\1\\0},\pmatrix{0\\0\\1}\right\}$

In my preliminery thinking, the span is the combination of vectors that indicate each vector in R, but in reverse, If I know a matrix, how can I derive the span?

Could anyone tell me I can I calculate the span of a matrix

Taking this as an example, how can I get to this result?

Answer 1

Your question is ambiguous, cause in general, for fixed $n,m$, the set $S=M_{n\times m}(\mathbb{K})$ (matrices of $n\times m$ with entries in the field $\mathbb{K}$) is a vector space over $\mathbb{K}$. Then, if $A\in S$, definition of $span(A)$ is the usual definition for span of a vector in $S$.

However, I suppose indeed in your problem you are asking for the column space spanned by a matrix. i.e. the space spanned for the columns of the matrix.

Then, you are searching for $V=span\{(1,1,0)^T,(1,1,2)^T,(1,1+2a,1+2b)^T\}$ and, by hypothesis, this is $span\{(1,1,0)^T,(0,0,1)^T\}$.

As vectors $(1,1,0)^T,(0,0,1)^T$ are linearly independents, you need that $(0,0,1)^T$ let a linear combination of $(1,1,2)^T,(1,1+2a,1+2b)^T$. That is, find $x,y$ such that

$$ \left\{\begin{array}{ccc}x+y&=&0\\x+y(1+2a)&=&0\\2x+y(1+2b)&=&1\end{array}\right. $$

for given $a,b$.

Solving for $x$ and $y$, we get, for $a\neq 0$, there are not $x,y$. For $a=0$, and $b=-3/2$, there are not solutions, and for $a=0$ and $b\neq -3/2$, $x=-1/(3+2b)$ and $y=1/(3+2b)$

Answer 2

You do not speak about span of matrices (in a vector space where matrix considered an abstract vector). In your case you should speak about column space $col(R)=Span\{R_{\downarrow j}\}_{j=1}^n$ or row space $row(R)=Span\{R_{i\rightarrow}\}_{i=1}^m$ of a matrix. There is also another related space, a null space, i.e. $Null(A)=\{x|Ax=0\}$.

The Span's argument, i.e. the set in the curly brackets may be reduced in case of the vectors, columns or rows respectively, are not linearly independent. More precisely you can remove any linearly dependent vector without changing the space this set spans.

Now to find the linearly independent vectors you simply produce with matrix reduction. For the rows you throw away the zero rows. For the columns only the pivot columns are linearly independent.

Answer 3

Observe that $\pmatrix{ 1 \\ 1 \\ 2} = 1\cdot \pmatrix{ 1 \\ 1 \\ 0} + 2\cdot \pmatrix{0 \\ 0 \\ 1}$, and $\pmatrix{1 \\ 1+ 2a\\ 2+ 2b} = 1\cdot \pmatrix{1 \\ 1 \\0}+(2+2b)\cdot \pmatrix{0 \\ 0 \\1} + 2a\cdot \pmatrix{0 \\ 1 \\0}$, and also $\text{det}\pmatrix{1 & 0 & 0 \\ 1 & 0 & 1\\ 0 & 1 & 0}= -1 \neq 0$, thus $R$ is spanned by $\pmatrix{1 \\ 1 \\ 0}$, $\pmatrix{0 \\ 0 \\1}$, and $\pmatrix{0 \\ 1 \\ 0}$.

Answer 4

It seems you're meant to find the values of $a$ and $b$ that give the column space $$\mathrm{col}(R)=\mathrm{span}\left\{\pmatrix{1\\1\\0},\pmatrix{0\\0\\1}\right\}=\left\{\pmatrix{r\\r\\s}:r,s \in \mathbb{R}\right\}.$$

We can immediately see that $a$ must be $0$, otherwise the third column of $R$ has different values in its first and second coordinate, which is not in the above span.

When $a=0$, $$R=\pmatrix{1&1&1\\1&1&1\\0&2&2+2b},$$ and all three columns of $R$ can be seen to belong to $\mathrm{col}(R)$ (set$r=1$ and $s=0$, $1$ or $2+2b$ accordingly), so $b$ can be any value in $\mathbb{R}$.