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Let $A = \begin{bmatrix} 1 & -1 & -5 & 1 & 4\\ -1 & 2 & 8 & -3 & -4\\ 3 & -1 & -9 & 0 & 4 \\ 2 & 2 & 2 & -5 & -10\\ 0&-3&-9&5&13\end{bmatrix}$

Now we define the subspace $W_{1},W_{2}$ of $A$ as follows -

$W_{1} = \{X \in M_{5 \times 5}| AX = 0\}$

$W_{2} = \{Y \in M_{5 \times 5} | YA =0\}$

I can see that $W_{1}$ is the nullspace of $A$

using rank nullity theorem I got Nullity of $A$ as $2$ since we have the rank of matrix $A$ to be 3.

Now I am thinking about the dimension of $W_{2}$?

As from the comments and we know that row rank = column rank, hence dim$(W_{2}) = 2$

But Now I am thinking about the dimension of $W_{1} \cap W_{2}$ and $W_{1} + W_{2}$?

Any ideas?

Martin Sleziak
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BAYMAX
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2 Answers2

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$W_1$ is not the nullspace of $A$. The nullspace of $A$ is a subspace of $\mathbb R^5$, while $W_1$ is a subspace of $\mathbb R^{5\times 5}$. To calculate the dimension of $W_1$, take into account that if the columns of $X$ are $[x_1,x_2,x_3,x_4,x_5]$ (where $x_i\in\mathbb R^5$ for all $i$), then $AX=[Ax_1,Ax_2,Ax_3,Ax_4,Ax_5]$.

With $W_2$, consider the fact that $YA=0\iff A^\top Y^\top=0$

5xum
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  • Each of $Ax_{i} = 0_{5 \times 1}$, so how can we conclude its dimension, it seems it has no independent columns and hence seems to have dimension 1? – BAYMAX Oct 24 '18 at 08:55
  • @BAYMAX I have no idea what you are asking me. I don't know what "it" is in the sentence "it has no independent columns". Literally, I don't know what your sentence means. – 5xum Oct 24 '18 at 08:57
  • Sorry, how the above representation of $AX$ by you can help us calculate dimension of $W_{1}$? – BAYMAX Oct 24 '18 at 08:58
  • @BAYMAX It allows you to see when exactly $AX=0$. $AX=0$ if and only if $Ax_i=0$ for all columns of $X$. Therefore, it is $0$ if and only if every column of $X$ is in the nullspace of $A$. – 5xum Oct 24 '18 at 08:59
  • Now if every column of $X$ is in the nullspace of $A$ then actually still i mam thinking how tht helps o calculate the dimension of $W_{1}$? – BAYMAX Oct 24 '18 at 09:06
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    @BAYMAX It helps you find the basis of $W_1$, from the basis of the nullspace of $A$. You still have some work ahead of you, but that's where you should start. And don't ask for help $5$ minutes after getting a tip. You can't work everything out in $5$ mintues. Math takes time. – 5xum Oct 24 '18 at 09:11
  • Sorry, i will keep in mind!, I got the dimension of the null space of $A$ to be 1. so if $x_{1}$ is the basis vector of null space of $A$, then remaining vectors $x_{2},x_{3},x_{4},x_{5}$ are the multiples of $x_{1}$.So i think that dimension of $W_{1} = 1$, am i correct?@5xum – BAYMAX Oct 24 '18 at 09:42
  • @BAYMAX Incorrect. Let $x$ be a non zero null vector of $A$. Then, both $[x,0,0,0,0]$ and $[0,x,0,0,0]$ are both elments of $W_1$, but the two matrices are linearly independent. – 5xum Oct 24 '18 at 09:48
  • Yes that is nice!!, So dimension of $W_{1}=5$, similarly dimension of null-space of $A^t$ is 1, thus similarly $W_{2}$ has dimension 5, but then $W_{1} \cap W_{2} =10$ as those 10 matrices are linearly independent?is this right? – BAYMAX Oct 24 '18 at 10:08
  • But i am thinking how can it be greater than 5? – BAYMAX Oct 24 '18 at 10:22
  • @BAYMAX Because $W_1$ is a subspace of a $25$-dimensional space... – 5xum Oct 24 '18 at 11:08
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    I agree with u completely!!, I am understanding as u asist me, would u mind to please look at the followup question here - https://math.stackexchange.com/questions/2968888/injectivity-and-subspaces-of-g-1-g-2-g-3 ? – BAYMAX Oct 24 '18 at 11:11
  • But that would mean that dim$(W_{1}+W_{2}) = 0= dim(W_{1})+dim(W_{2}) - dim(W_{1} \cap W_{2})$ how can i think of dimension zero here @5xum? – BAYMAX Oct 24 '18 at 12:17
  • Why would $\dim(W_1+W_2)=0$ be true? $W_1$ is a nonzero space, therefore, $W_1\subseteq W_1+W_2$, which means that $\dim(W_1+W_2)\geq \dim(W_1)>0$! – 5xum Oct 24 '18 at 12:21
  • But we have the formula $dim(W_{1} +W_{2}) = dim(W_{1}) +dim(W_{2}) - dim(W_{1} \cap W_{2})$ ? is that incorrect? and we have $dim(W_{1}) =dim(W_{2}) = 5,dim(W_{1} \cap W_{2}) = 10$ – BAYMAX Oct 24 '18 at 12:23
  • @BAYMAX That formula is correct. But I don't see why you think $\dim(W_1\cap W_2)=10$... – 5xum Oct 24 '18 at 12:24
  • I thought it was correct as i mentioned in my comments a bit far above, can u please help how should i think about $dim(W_{1} \cap W_{2})$ ? number of linearly independent matrices $X$ such that bith $AX=0$ and $XA=0$ – BAYMAX Oct 24 '18 at 12:25
  • @BAYMAX I now see you wrote it in the comments, but it's false. I don't know why you would think $\dim W_1\cap W_2)=10$, since $W_1\cap W_2\subseteq W_1$. To calculate the dimension of $W_1\cap W_2$, I see no other way apart from actually calculating a basis for $W_1$ and then seeing which of the basis elements are in $W_2$. – 5xum Oct 24 '18 at 12:28
  • We get 5 linearly inependent matrix from $AX=0$ and another $5$ linearly independent independent matrices from $YA=0$, then what should we do? – BAYMAX Oct 24 '18 at 12:32
  • @BAYMAX From reading your comments, it appear that you might be confusing the columns of $X\in W_1$ being independent with the matrices of $W_1$ being independent. These are not the same concept. Hint: $\dim W_1$ is larger than $5$. – Michael Burr Oct 25 '18 at 11:36
  • @MichaelBurr oh i see, but i was thinking the same thing as you pointed out wrong? how could i think about the dimension of $W_{1} >5$ – BAYMAX Oct 25 '18 at 12:01
  • @BAYMAX 5xum has already hinted at a basis: think about matrices of the form $[x,0,0,0,0]$ or $[0,x,0,0,0]$ or $[0,0,x,0,0]$ or ... where $x$ is a vector from the null space of $A$ and $0$ is the zero vector. Are these independent? How many of these are there? Do they span (think about linearity and different ways of writing matrix multiplication)? – Michael Burr Oct 25 '18 at 13:12
  • @MichaelBurr yes, so since there are $5$ linearly independent matrices( null space of $A$ being $1$), so the required dimension of $W_1$ should be $5$ right? – vidyarthi Oct 26 '18 at 06:33
  • In your question, you state that the null space is two dimensional. So, is it one or two dimensional? – Michael Burr Oct 26 '18 at 10:13
2

Let us first look at \begin{align*} W'&= \{X\in M_{5\times 1}; AX=0\}\\ W''&= \{Y\in M_{1\times 5}; YA=0\} \end{align*} In the other words we look at similar equations but with column/row vectors instead of matrices.

By a direct computation you can get that $\operatorname{rank}A=4$, which implies that $\dim(W')=\dim(W'')=1$. You can also compute that $W'$ is the span of the column vector $\vec a=(2,-3,1,0,0)^T$ and that $W''$ is the span of the row vector $\vec b^T=(5,0,-1,-1,2)$.

If we denote the columns of the matrix $X$ as $\vec c_1,\dots,\vec c_5$ then we have $$AX = A\begin{pmatrix} \vec c_1 & \vec c_2 & \ldots & \vec{c_5} \end{pmatrix} = \begin{pmatrix} A\vec c_1 & A\vec c_2 & \ldots & A\vec{c_5} \end{pmatrix} = \begin{pmatrix} \vec 0 & \vec 0 & \ldots & \vec 0 \end{pmatrix}.$$ I.e., each of the columns fulfills the condition $A\vec c_i=\vec 0$. So we see that the matrices in $W_1$ are precisely those matrices where each column is a multiple of $\vec a$.

Similarly, we get for the rows of the matrix $X\in W''$ the condition $\vec r_i^TA=\vec 0^T$, and $W_2$ consists of those matrices where each row is multiple of $\vec b$.

We get that \begin{align*} W_1&=\{ \begin{pmatrix} 2a & 2b & 2c & 2d & 2e \\ -3a & -3b & -3c & -3d & -3e \\ a & b & c & d & e \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}; a,b,c,d,e \in \mathbb R\} \\ W_2&=\{ \begin{pmatrix} 5s & 0 & -s & -s & -2s \\ 5t & 0 & -t & -t & -2t \\ 5u & 0 & -u & -u & -2u \\ 5v & 0 & -v & -v & -2v \\ 5w & 0 & -w & -w & -2w \\ \end{pmatrix}; s,t,u,v,w \in \mathbb R\} \end{align*} And we also see that $\dim(W_1)=\dim(W_2)=5$.

Now the matrices in the intersection $W_1\cap W_2$ are precisely the matrices which can be expressed in both ways.

$$\begin{pmatrix} 2a & 2b & 2c & 2d & 2e \\ -3a & -3b & -3c & -3d & -3e \\ a & b & c & d & e \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}= \begin{pmatrix} 5s & 0 & -s & -s & -2s \\ 5t & 0 & -t & -t & -2t \\ 5u & 0 & -u & -u & -2u \\ 5v & 0 & -v & -v & -2v \\ 5w & 0 & -w & -w & -2w \\ \end{pmatrix} $$ Those are precisely the multiples of $$ \begin{pmatrix} 10& 0 &-2 &-2 &-4 \\ -15& 0 & 3 & 3 & 6 \\ 5 & 0 &-1 &-1 &-2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} $$ This matrix generates $W_1\cap W_2$. We see that $\dim(W_1\cap W_2)=1$.

From the equation $$\dim W_1+\dim W_2=\dim(W_1+W_2)+\dim(W_1\cap W_2)$$ we can calculate that $\dim(W_1+W_2)=9$.

Martin Sleziak
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