[SOLVED] Basis of the nullspace and image of a function

**bmp05** · May 21st 2009, 03:34 AM

Let $f: M_{22} (\mathbb{R}) \rightarrow \mathbb{R}[T]$ defined by $f \left[\begin{matrix} a & b\\ c & d \end{matrix}\right] = (a+b) + (a + b)T + (a + b + c + d)T^2$ for all $\left[\begin{matrix} a & b\\ c & d \end{matrix}\right] \in M_{22}$ .

1. Prove that $f$ is linear.
2. Calculate the basis of the null-space of $f$ and of the image of $f$ .

How do I even prove that $f$ is linear?

**NonCommAlg** · May 21st 2009, 04:24 AM

Originally Posted by bmp05

Let $f: M_{22} (\mathbb{R}) \rightarrow \mathbb{R}[T]$ defined by $f \left[\begin{matrix} a & b\\ c & d \end{matrix}\right] = (a+b) + (a + b)T + (a + b + c + d)T^2$ for all $\left[\begin{matrix} a & b\\ c & d \end{matrix}\right] \in M_{22}$ .

1. Prove that $f$ is linear.
2. Calculate the basis of the null-space of $f$ and of the image of $f$ .

How do I even prove that $f$ is linear?

for 1) you need to show that for any matrices $A,B \in M_2(\mathbb{R}), \ \alpha \in \mathbb{R}: \ f(\alpha A + B)=\alpha f(A) + f(B).$ this is straightforward.

for 2): $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ is in the null-space of $f$ if and only if $a+b=a+b+c+d=0,$ which gives us: $b=-a, \ d=-c.$ thus:

$A=\begin{bmatrix} a & -a\\ c & -c \end{bmatrix}=a \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}+ c \begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix}.$ so a basis for the null-space of $f$ is ... ?

**bmp05** · May 21st 2009, 04:42 AM

Thanks NonCommAlg!

So to prove that f is linear you have to show that the elements in f are scalar multiples of each other? As in $y = mx$

And the null space are derived by taking the coefficients of f so that: $(a+b) + (a + b)T + (a + b + c + d)T^2 = 0$

Wow. That really did help. Sometimes, I feel like stunned rabbit when confronted with new terminology/ presentation of a question

. Thank you very very much!

Can you recommend a good text book, for linear algebra?

**NonCommAlg** · May 21st 2009, 05:00 AM

Originally Posted by bmp05

Thanks NonCommAlg!

So to prove that f is linear you have to show that the elements in f are scalar multiples of each other? As in $y = mx$

that's not what i said!

i said choose any two matrices in $M_2(\mathbb{R}): \ A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \ B=\begin{bmatrix} a' & b'\\ c' & d' \end{bmatrix},$ and a scalar $\alpha \in \mathbb{R}.$ then:

$\alpha A + B=\begin{bmatrix} \alpha a + a' & \alpha b + b' \\ \alpha c + c'& \alpha d + d'\end{bmatrix}.$ then from the definition of $f$ we have: $f(A)=a+b+(a+b)T + (a+b+c+d)T^2$ and

$f(B)=a'+b'+(a'+b')T + (a'+b'+c'+d')T^2.$ now you have everything ready to find both $\alpha f(A) + f(B)$ and $f(\alpha A + B)$

and see that they're equal, which exactly means that $f$ is linear.

**bmp05** · May 21st 2009, 05:59 AM

Ok- well, what does linearity mean, exactly? Is there anything special about linearity? Are there vector-spaces which are for example not linear?

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