Finding Basis of Null Space and Range

[Gooolati]

Homework Statement

Prove T is a linear transformation and find bases for both N(T) and R(T).

Homework Equations

The Attempt at a Solution

T:M_2x3(F) \rightarrow M_2x2(F) defined by:

T(a₁₁ a₁₂ a₁₃)
(a₂₁ a₂₂ a₂₃)

(this is one matrix)
=

(2a₁₁-a₁₂ a₁₃+2a₁₂)
( 0 0)

(this is one matrix)

So I verified that it is a linear transformation by checking that T(cx+y)=cT(x)+T(y). But I don't understand how to find a basis for the null space and range.

I can see that since N(T)={x:T(x)=0} that N(T) here it all vectors of the form:

(t/2 t -2t)
( b b b)

(this is one matrix)

Since the 2nd row in our domain always goes to 0, the second row is arbitrary, which I represented by b.

How do I find a basis for all multiples of the matrix

t(1/2 1 -2)
( b b b)?

And I'm not even sure on how to start off finding the basis for the range. All help is appreciated. Thanks!

 

[Deveno]

do you have any idea of what the dimension of M_2x3(F) is?

can you think of a basis for it?

can you express N(T) in terms of this basis?

alternately: do you know how to "vectorize" a matrix?

suppose you "vectorized" the elements of M_2x3(F) and M_2x2(F). could you pick bases for these spaces, and calculate a matrix for T in terms of those bases? what size would it have? could you find its rank and nullity?

 

[Gooolati]

do you have any idea of what the dimension of M2x3(F) is?

Since it is 2x3, the dimension would be 6?

Because we would need 6 vectors:

\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \end{array},

\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \end{array},

\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \end{array}

\begin{array}{ccc}
0 & 0 & 0 \\
0 & 1 & 0 \end{array},

\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \end{array}

\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \end{array}

N(T) in terms of this basis would be multiples of:

(1/2)(first matrix)+(third matrix)+(-2)(fifth matrix)?

Something doesn't seem right to be hmm.

Also, what do you mean by vectorizing?

 

[vela]

do you have any idea of what the dimension of M2x3(F) is?

Since it is 2x3, the dimension would be 6?

Because we would need 6 vectors:

<snip>

N(T) in terms of this basis would be multiples of:

(1/2)(first matrix)+(third matrix)+(-2)(fifth matrix)?

Something doesn't seem right to be hmm.

Good start. But what about the second matrix? Doesn't T map that to 0 as well? What about the other basis elements?

Also, what do you mean by vectorizing?

You can express any matrix in M_2x3 as a linear combination of those six matrices, so if you label those six matrices e₁, … , e₆, you could write
$$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} = a_{11}\mathbf{e}_1 + a_{21}\mathbf{e}_2 + a_{12}\mathbf{e}_3 + a_{22}\mathbf{e}_4 + a_{13}\mathbf{e}_5 + a_{23}\mathbf{e}_6$$ This matrix would therefore correspond to the coordinate vector $(a_{11}, a_{21}, a_{12}, a_{22}, a_{13}, a_{23})^T$.