# Injective, surjective map!

• Dec 13th 2010, 06:55 AM
cantona
Injective, surjective map!
Hello everyone!

May anyone can help in this case.

I have a map F: $\left F( \begin{bmatrix}
d_1_1 & d_1_2 & \ d_1_3 \\
d_2_1 & d_2_2 & d_2_3 \\
d_3_1 & d_3_2 & d_3_3
\end{bmatrix}) \right = \begin{bmatrix}
d_1_1 & d_1_2-d_2_1 & d_3_1 \\
0 & 2d_2_2 & d_2_1-d_1_2 \\
0 & 0 & 3d_3_3
\end{bmatrix}$

in a different note:

$\left F( \begin{bmatrix}
d_1_1 & d_1_2 & \ d_1_3 \\
d_2_1 & d_2_2 & d_2_3 \\
d_3_1 & d_3_2 & d_3_3
\end{bmatrix}) \right = \begin{bmatrix}
g_1_1 & g_1_2 & g_1_3 \\
0 & g_2_2 & g_2_3 \\
0 & 0 & g_3_3
\end{bmatrix}$

eg. $\left F( \begin{bmatrix}
0 & 3 & \ 6 \\
1 & 4 & 7 \\
2 & 5 & 8
\end{bmatrix}) \right = \begin{bmatrix}
0 & 2 & 2 \\
0 & 8 & -2 \\
0 & 0 & 24
\end{bmatrix}$
.

I must to find out if this map is linear, surjective, injective?

The first point I already done, so I know the map is linear. I have more problems with second and third point. My thinking:
1) the map is not injective because:

$for \ d_2_1 \ne d_3_1 \ is \ F(d_2_1)=F(d_3_1) \ or \ g_2_1=g_3_1=0$

2) but the map is surjective because:
for every g in G exits d in D that is F(d)=g

Are those two arguments correct?

Enrico
• Dec 13th 2010, 07:46 AM
Drexel28
Quote:

Originally Posted by cantona
Hello everyone!

May anyone can help in this case.

I have a map F: $\left F( \begin{bmatrix}
d_1_1 & d_1_2 & \ d_1_3 \\
d_2_1 & d_2_2 & d_2_3 \\
d_3_1 & d_3_2 & d_3_3
\end{bmatrix}) \right = \begin{bmatrix}
d_1_1 & d_1_2-d_2_1 & d_3_1 \\
0 & 2d_2_2 & d_2_1-d_1_2 \\
0 & 0 & 3d_3_3
\end{bmatrix}$

in a different note:

$\left F( \begin{bmatrix}
d_1_1 & d_1_2 & \ d_1_3 \\
d_2_1 & d_2_2 & d_2_3 \\
d_3_1 & d_3_2 & d_3_3
\end{bmatrix}) \right = \begin{bmatrix}
g_1_1 & g_1_2 & g_1_3 \\
0 & g_2_2 & g_2_3 \\
0 & 0 & g_3_3
\end{bmatrix}$

eg. $\left F( \begin{bmatrix}
0 & 3 & \ 6 \\
1 & 4 & 7 \\
2 & 5 & 8
\end{bmatrix}) \right = \begin{bmatrix}
0 & 2 & 2 \\
0 & 8 & -2 \\
0 & 0 & 24
\end{bmatrix}$
.

I must to find out if this map is linear, surjective, injective?

The first point I already done, so I know the map is linear. I have more problems with second and third point. My thinking:
1) the map is not injective because:

$for \ d_2_1 \ne d_3_1 \ is \ F(d_2_1)=F(d_3_1) \ or \ g_2_1=g_3_1=0$

2) but the map is surjective because:
for every g in G exits d in D that is F(d)=g

Are those two arguments correct?

Enrico

For injectivitgy you need to give specific numbers for which this isn't true. Also, assuming this is a map from $3\times 3$ matrices over a field to itself then a linear map is injective if and only if it's surjective, so keep this in mind.
• Dec 13th 2010, 08:31 AM
cantona
Quote:

Originally Posted by Drexel28
For injectivitgy you need to give specific numbers for which this isn't true. Also, assuming this is a map from $3\times 3$ matrices over a field to itself then a linear map is injective if and only if it's surjective, so keep this in mind.

What you mean with this specific numbers?

Maybe that:

$if \ d_2_1 = d_3_1 =d_3_2 \ and \ F(d_2_1)=F(d_3_1)=F(d_3_2)=0$ there is a lot of possibility that map is injective. Or the case of null matrix.
• Dec 13th 2010, 08:35 AM
Drexel28
Quote:

Originally Posted by cantona

What you mean with this specific numbers?

Maybe that:

$if \ d_2_1 = d_3_1 =d_3_2 \ and \ F(d_2_1)=F(d_3_1)=F(d_3_2)=0$ there is a lot of possibility that map is injective. Or the case of null matrix.

I mean that you need to exhibit a specific (or a class of specific) non-zero matrices which map to zero. So far you've tried applying a function of matrices to scalars.
• Dec 13th 2010, 09:09 AM
cantona
Quote:

Originally Posted by Drexel28
I mean that you need to exhibit a specific (or a class of specific) non-zero matrices which map to zero. So far you've tried applying a function of matrices to scalars.

If I understand you mean this:
if D is
$\ d_1_1=d_2_2=d_3_1=d_3_3=0 \ and \ d_1_2=d_2_1 \ and\ d_1_3,d_2_3,d_3_2 \in \mathbb{R}$
G is null matrix and then the map is not injective. If I add this note:

$for \ d_2_1 \ne d_3_1 \ne d_3_2 \ is \ F(d_2_1)=F(d_3_1)=F(d_3_2) \ or \ g_2_1=g_3_1=g_3_2=0$

Is this enough to prove that map is not injective?
• Dec 13th 2010, 09:36 AM
snowtea
Be careful about the definitions of injective and surjective. As Drexel28 mentioned, this is asking about the injectivity and surjectivity of F (which maps 3 x 3 matrices to other 3 x 3 matrices). Your domain and range are 3 x 3 matrices.

$F : M_{3\times 3} \to M_{3\times 3}$

Let $M_1, M_2$ be 3 x 3 matrices.

Here's how injectivity and surjectivity should be interpreted in the problem:
Injective means
$\forall M_1, M_2 : F(M_1) = F(M_2) \Rightarrow M_1 = M_2$

Surjective means
$\forall M_1, \exists M_2 : F(M_2) = M_1$

The entire time, we are talking about the entire matrix, not the elements inside the matrix.
• Dec 13th 2010, 11:59 AM
cantona
Thank you.

So if we have two matrices:

$A=\begin{bmatrix}
0 & 1 & 4 \\
1 & 0 & 3 \\
0 & 2 & 0
\end{bmatrix} \right B= \begin{bmatrix}
0 & 3 & 7 \\
3 & 0 & 5 \\
0 & 1 & 0
\end{bmatrix}$

$F(A)=F(B)=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$

So the map is not injective because $F(A)=F(B) \ but \ A\ne B$.

If we change only elements $d_1_3,d_2_3,d_3_2$ and all the others stay the same we always get the same output matrices for those set. Special case is when we get the null matrix, then must be:

$\ d_1_1=d_2_2=d_3_1=d_3_3=0 \ ; \ d_1_2=d_2_1 \ ; \ d_1_3,d_2_3,d_3_2 \in \mathbb{R}$

Is that right?
• Dec 13th 2010, 12:08 PM
snowtea
Yes, so you have shown it is not injective.
You did not even need the second part about $d_{13}, d_{23}, d_{32}$, but it is a good observation.

For the second part, is it surjective?

Hint: Can you find an A s.t.
$
F(A)=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix}
$

?
• Dec 13th 2010, 12:17 PM
Drexel28
Quote:

Originally Posted by snowtea
Yes, so you have shown it is not injective.
You did not even need the second part about $d_{13}, d_{23}, d_{32}$, but it is a good observation.

For the second part, is it surjective?

Hint: Can you find an A s.t.
$
F(A)=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix}
$

?

In fact, he didn't even have to show that $F(B)=F(A)$ since by showing that $F(A)=[0]$ he showed that $\ker F$ is non-trivial and thus $F$ can't be injective.

For surjectivity your hint is a very good one, since in this case it's clear what the result is but you could have (if you are aware of it) appealed to the (common?) theorem which says that if $V$ is finite dimensional then $T:V\to V$ is a monomorphism (fancy word for injective linear transformation) if and only if it's an epimorphism (fancy word for surjective linear transformation). This is just a comment though.
• Dec 13th 2010, 01:28 PM
cantona
I do not understand this hint because in my case there is no chance to get such output after mapping.
• Dec 13th 2010, 01:28 PM
snowtea
• Dec 13th 2010, 01:56 PM
cantona
So if I note this in more formal form:

We take the matrix $M_1=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix}$

In our case the map not allows us to get such output so the form that provides surjectivity does not stand up.

So $F(M_2) \ne M_1$ and the map is not surjective.

So all of upper triangular matrices are not surjective?
• Dec 13th 2010, 02:13 PM
snowtea
Yes, if F(M) is always upper triangular, then it cannot be surjective.
Because the mapping "misses" all matrices that are not upper triangular.
• Dec 14th 2010, 12:49 AM
cantona
Thank you both for your help.