- Thread starter
- #1

M = {a b

0 d}

where a, b and d are complex.

I've found a basis for M but need to know how to find the set of scalar homomorphisms of M from these.

I have the basis as

M_1 = {1 0

0 1}

M_2 = {0 1

0 0}

and

M_3 = {0 0

0 1}

Any ideas?

- Thread starter dray
- Start date

- Thread starter
- #1

M = {a b

0 d}

where a, b and d are complex.

I've found a basis for M but need to know how to find the set of scalar homomorphisms of M from these.

I have the basis as

M_1 = {1 0

0 1}

M_2 = {0 1

0 0}

and

M_3 = {0 0

0 1}

Any ideas?

- Jan 29, 2012

- 661

$A_1=\begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix}$, $A_2=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}$, $A_3=\begin{bmatrix}{0}&{0}\\{0}&{1}\end{bmatrix}$

The coordinates of $A=\begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix}$ with respect to $B_M$ are $(a,b,d)^t$ so we can express the set of all scalar homomorphisms of $M$ in the matricial form:

$f_\lambda \begin{bmatrix}{a}\\{b}\\{d}\end{bmatrix}=\lambda I_3 \begin{bmatrix}{a}\\{b}\\{d}\end{bmatrix}\quad (\lambda\in\mathbb{C})$

- Moderator
- #3

- Feb 7, 2012

- 2,799

It is not entirely clear from the question, but I think that you are asking how to determine all the ring (or algebra) homomorphisms from the set $M$ to the scalars. Notice that two of your three basis matrices are idempotent: $M_1^2=M_1$ and $M_3^2=M_3.$ If $f:M\to\mathbb{C}$ is a homomorphism, it follows that $\bigl(f(M_1)\bigr)^2 = f(M_1)$ and hence $f(M_1)$ must be 0 or 1. Similarly $f(M_3)$ must be 0 or 1.Let M be the set of 2x2 matrices defined by

M = a b

0 d}

where a, b and d are complex.

I've found a basis for M but need to know how to find the set of scalar homomorphisms of M from these.

I have the basis as

M_1 = {1 0

0 1}

M_2 = {0 1

0 0}

and

M_3 = {0 0

0 1}

Any ideas?

Next, $M_1+M_3=I$ (the identity matrix), so $f(M_1)+f(M_3) = f(I) = 1$ (unless $f$ is the identically zero map). Thus if $f(M_1)=1$ then $f(M_3)=0$ and vice versa.

Use those facts to show that the only two homomorphisms from $M$ to the scalars are $\begin{bmatrix}a&b \\ 0&d \end{bmatrix} \mapsto a$ and $\begin{bmatrix}a&b \\ 0&d \end{bmatrix} \mapsto d.$

- Thread starter
- #4

Thanks for this Opalg.It is not entirely clear from the question, but I think that you are asking how to determine all the ring (or algebra) homomorphisms from the set $M$ to the scalars. Notice that two of your three basis matrices are idempotent: $M_1^2=M_1$ and $M_3^2=M_3.$ If $f:M\to\mathbb{C}$ is a homomorphism, it follows that $\bigl(f(M_1)\bigr)^2 = f(M_1)$ and hence $f(M_1)$ must be 0 or 1. Similarly $f(M_3)$ must be 0 or 1.

Next, $M_1+M_3=I$ (the identity matrix), so $f(M_1)+f(M_3) = f(I) = 1$ (unless $f$ is the identically zero map). Thus if $f(M_1)=1$ then $f(M_3)=0$ and vice versa.

Use those facts to show that the only two homomorphisms from $M$ to the scalars are $\begin{bmatrix}a&b \\ 0&d \end{bmatrix} \mapsto a$ and $\begin{bmatrix}a&b \\ 0&d \end{bmatrix} \mapsto d.$

I can't see how $M_1+M_3=I$ for the basis matrices I have determined. The question I am tackling, is from Maddox and he specifically requests that this basis be found and then used to find the set of all scalar homomorphisms of the set M.

- Moderator
- #5

- Feb 7, 2012

- 2,799

$M_1+M_3 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$.I can't see how $M_1+M_3=I$ for the basis matrices I have determined.

- Thread starter
- #6

But my bases (which are the ones requested in the question) are$M_1+M_3 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$.

$M_1=\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

$M_2=\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$

$M_3=\begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

- Moderator
- #7

- Feb 7, 2012

- 2,799

So that was just me misreading the question as usual.But my bases (which are the ones requested in the question) are

$M_1=\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

$M_2=\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$

$M_3=\begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

Of course, the matrix that I called $M_1$ is actually $M_1-M_3$. If you repeat my solution to the problem, replacing my $M_1$ by $M_1-M_3$, then you should have a valid solution in terms of the basis requested in the question.

- Thread starter
- #8

Thanks. That now makes more sense.So that was just me misreading the question as usual.

Of course, the matrix that I called $M_1$ is actually $M_1-M_3$. If you repeat my solution to the problem, replacing my $M_1$ by $M_1-M_3$, then you should have a valid solution in terms of the basis requested in the question.

Can I ask how you got the two scalar homomorphisms that map to $a$ and $d$ respectively? I need to use these to show that the radical of $M$ is $\begin{bmatrix}0&b\\0&0\end{bmatrix}$, where $b\in\mathbb{C}$.

- Moderator
- #9

- Feb 7, 2012

- 2,799

You are looking for maps $f:M\to\mathbb{C}$ such that $f(A_1A_2) = f(A_1)f(A_2)$ for all $A_1,A_2\in M.$ If $A_1 = \begin{bmatrix}a_1&b_1 \\ 0&d_1 \end{bmatrix}$ and $A_2 = \begin{bmatrix}a_2&b_2 \\ 0&d_2 \end{bmatrix}$ then $A_1A_2 = \begin{bmatrix}a_1a_2&a_1b_2+b_1d_2 \\ 0&d_1d_2 \end{bmatrix}.$ If you stare at that equation for a while, you should notice that the maps taking the matrix to its top left or bottom right elements preserve multiplication. After a bit more calculation you can see that these maps also preserve inverses and are therefore multiplicative homomorphisms.Can I ask how you got the two scalar homomorphisms that map to $a$ and $d$ respectively?

- Thread starter
- #10

Thanks for this. Although I an now a little confused as to why we wanted to find the numbers for $f(M_1)$ and so on. How does knowing this enable us to find the scalar homomorphism $f$?You are looking for maps $f:M\to\mathbb{C}$ such that $f(A_1A_2) = f(A_1)f(A_2)$ for all $A_1,A_2\in M.$ If $A_1 = \begin{bmatrix}a_1&b_1 \\ 0&d_1 \end{bmatrix}$ and $A_2 = \begin{bmatrix}a_2&b_2 \\ 0&d_2 \end{bmatrix}$ then $A_1A_2 = \begin{bmatrix}a_1a_2&a_1b_2+b_1d_2 \\ 0&d_1d_2 \end{bmatrix}.$ If you stare at that equation for a while, you should notice that the maps taking the matrix to its top left or bottom right elements preserve multiplication. After a bit more calculation you can see that these maps also preserve inverses and are therefore multiplicative homomorphisms.