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Homomorphisms and radical

dray

Member
Feb 5, 2012
37
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?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
A more natural basis is $B_M=\{A_1,A_2,A_3\}$ with

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

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:

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
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?
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 $\begin{bmatrix}a&b \\ 0&d \end{bmatrix} \mapsto d.$
 

dray

Member
Feb 5, 2012
37
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 $\begin{bmatrix}a&b \\ 0&d \end{bmatrix} \mapsto d.$
Thanks for this Opalg.

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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
I can't see how $M_1+M_3=I$ for the basis matrices I have determined.
$M_1+M_3 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$.
 

dray

Member
Feb 5, 2012
37
$M_1+M_3 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$.
But my bases (which are the ones requested in the question) are

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

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

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
But my bases (which are the ones requested in the question) are

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

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

$M_3=\begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$
So that was just me misreading the question as usual. (Blush)

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.
 

dray

Member
Feb 5, 2012
37
So that was just me misreading the question as usual. (Blush)

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.
Thanks. That now makes more sense.

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 $b\in\mathbb{C}$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
Can I ask how you got the two scalar homomorphisms that map to $a$ and $d$ respectively?
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.
 

dray

Member
Feb 5, 2012
37
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.
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 $f$?