# Math Help - More Confusion, (linear algebra)

1. ## More Confusion, (linear algebra)

Handing in date is drawing very close and im losing my mind over this.

\begin{pmatrix}1 & 4 & 6\\3 & 9 & 1\\2 & 6 & 5\\4 & 4 & 9\end{pmatrix}

ive been asked to find the rref(M) which is confusing me because I just end up with:

\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}

which kind of makes sence to me as there are 3 variables so one of the rows has to be a linear combination of the others? but im then asked to find the kernal(M) which seems to me to be the zero vector????? how confusing...

2. Originally Posted by SilenceInShadows
Handing in date is drawing very close and im losing my mind over this.

\begin{pmatrix}1 & 4 & 6\\3 & 9 & 1\\2 & 6 & 5\\4 & 4 & 9\end{pmatrix}

ive been asked to find the rref(M) which is confusing me because I just end up with:

\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}

which kind of makes sence to me as there are 3 variables so one of the rows has to be a linear combination of the others? but im then asked to find the kernal(M) which seems to me to be the zero vector????? how confusing...
Your transform is $T:\mathbb{R}^3 \to \mathbb{R}^4$

Remember that the Kernal of a linear transformation is the set of vectors $v \in \mathbb{R}^3$ such that $T(v)=0$ where 0 is the 0 vector in $\mathbb{R}^4$

Solving this gives

$
\begin{pmatrix}1 & 4 & 6 & 0\\3 & 9 & 1 & 0\\2 & 6 & 5 & 0\\4 & 4 & 9 & 0\end{pmatrix}
$

$\begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\end{pmatrix}$

So the only solution to this system as you stated is the trivial solution. So the only vector in the kernel is $(0,0,0)$

Note: this tells us that the Linear transformation is 1-1.

Good luck.

3. Thankyou,

I wanted the answer before I added this bit; I need a basis for the kernal.

I need to find a basis for the trival zero vector? hence my confusion...

4. also what would be the dimension of the null space? the rank + nullity gives the number of columns right? the rank is given by the numbr of linearly independent rows or columns, giving 3.

3 + nullity = 3?

So my matrix with kernal the zero vector that I have to find a basis for has dimension zero?

fun!

5. Originally Posted by SilenceInShadows
also what would be the dimension of the null space? the rank + nullity gives the number of columns right? the rank is given by the numbr of linearly independent rows or columns, giving 3.

3 + nullity = 3?

So my matrix with kernal the zero vector that I have to find a basis for has dimension zero?

fun!
The Dimenstion of a finite-dimensional vector space V is defined to be the number of vectors in a basis for V. In addition the zero vector space is defined to have dimension zero.

See examples in this wiki

Dimension (vector space) - Wikipedia, the free encyclopedia

Good luck.