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I have looked all through my book to figure out how to solve this type of problem and can't seem to find it anywhere.

For T(M)=$$T(M)=\begin{matrix}1&2\\0&3 \end{matrix}M$$ If no basis is specified use the standard basis
$\Upsilon=(1,t,t^2) $ for P2 $\Upsilon = $$\begin{matrix} |1&0|\\|0&0| \end{matrix}$$,$$\begin{matrix} |0&1|\\|0&0| \end{matrix}$,$\begin{matrix} |0&0|\\|1&0| \end{matrix}$,$\begin{matrix} |0&0|\\|0&1| \end{matrix}$

for $\mathbb{R}$ 2x2 and $\Upsilon=(1,i)$ for $\mathbb{C}$ for the subspace U 2x2 of upper triangular 2x2 matrices use a basis (from there it reiterates the given basis without the third matrix). Determine whether T is an isomorphism, find bases of the karnel and image of T, and thus determine the rank of T.

A sample problem that I already have the answer to is:

$$T(M)=\begin{matrix} |1&2|\\|0&3| \end{matrix}M$$ from U 2x2 to U 2x2.

The answer is $$\begin{matrix} |1&0&0|\\|0&1&2|\\|0&0&3| \end{matrix}$$

Unfortunately I don't really know where to begin. Maybe multiply the given matrix by each base? I don't really know where to go from there though in order to get an output for something that is a 3x3 matrix. In addition, given that the only examples I can find for isomorphisms are with like dimensioned matrices I don't really know how to tackle that part of the problem given the difference between the dimensions of the input and output.I do however know how to find kernel and image. This problem isn't one of the ones from my assignment, but is in the same section so if I can understand this the rest should come naturally. Any help would be greatly appreciated.

EDIT: I assume the 3x3 matrix given is the solution for T to the problem and not the confirmation of isomorphism or the rank/kernel solutions. Thanks to the commenter who pointed that out. Also, the symbol it gives in my book isn't actually an upsilon but a U in a strange gothic font. That being the case I'm assuming: $\Upsilon = U$

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  • $\begingroup$ Great latex start! The easiest way to make matrices in tex is to use \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}. Here, the pmatrix stands for "parentheses matrix", and it makes a matrix with parentheses. The & signs indicate different matrix entries, and the double backslash indicate that it's time to start the next row. So my example is an identity matrix. $\endgroup$
    – davidlowryduda
    Oct 30, 2015 at 19:56
  • $\begingroup$ I'd actually say the easiest way is \pmatrix{ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 } as it doesn't require the begin and end and renders exactly the same. Your way generalizes to other types of bracketing for matrices though: bmatrix begin square brackets, vmatrix being straight brackets, etc. $\endgroup$
    – user137731
    Oct 30, 2015 at 20:04

1 Answer 1

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Let's find out how they got $\pmatrix{1& 0 & 0 \\0 & 1 & 2 \\ 0 & 0 & 3}$.

Let's look at the transformation $T$ on $U_{2\times 2}$:

$$Tu = \pmatrix{1 & 2 \\ 0 & 3}\pmatrix{a & b \\ 0 & c} = \pmatrix{a & b+2c \\ 0 & 3c}$$

What's a good basis for $U_{2\times 2}$? Well let's just consider the arbitrary upper triangle matrix $$\pmatrix{a & b \\ 0 & c} = a\pmatrix{1 & 0 \\ 0 & 0} + b\pmatrix{0 & 1 \\ 0 & 0} + c\pmatrix{0 & 0 \\ 0 & 1}$$ This shows us that $$\left\{\pmatrix{1 & 0 \\ 0 & 0}, \pmatrix{0 & 1 \\ 0 & 0}, \pmatrix{0 & 0 \\ 0 & 1}\right\}$$ is a basis for $U_{2\times 2}$.

So in terms of this basis, the matrix $\pmatrix{a & b \\ 0 & c}$ has coordinate vector $\pmatrix{a \\ b \\ c}$ and the image of this matrix has coordinate vector $\pmatrix{a \\ b+2c \\ 3c}$.

So let's now figure out which matrix represents this transformation on these coordinate vectors:

$$\pmatrix{? & ? & ? \\ ? & ? & ? \\ ? & ? & ?}\pmatrix{a \\ b \\ c} = \pmatrix{a \\ b+2c \\ 3c} = \pmatrix{1a + 0b + 0c \\ 0a + 1b+2c \\ 0a + 0b + 3c}$$

Clearly the matrix which does this is $$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 3}$$

Now can you see where to go from here to solve the rest of the questions?

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  • $\begingroup$ That's awesome! I don't quite understand how you put the elements from the matrix product Tu equated to into a coordinate vector, but I think I understand the general concept. How would I determine if this was isomorphic? Wouldn't they be isomorphic if the 3x3 matrix we found could be multiplied by another matrix to be the identity matrix? $\endgroup$
    – tvengle
    Oct 30, 2015 at 20:31
  • $\begingroup$ For the coordinate vector part: what is a coordinate vector? It's just the coordinates of a "vector" (in some abstract vector space) put into a column matrix. The coordinates of the matrix $\pmatrix{a & b \\ 0 & c}$ are just $(a,b,c)^T$ with respect to the standard basis of $U_{2\times 2}$ (coordinates are just the coefficients on the basis expansion of the vector -- look at the expansion in my answer above). For the isomorphism: exactly. So just determine if that matrix is invertible (maybe by checking the determinant). $\endgroup$
    – user137731
    Oct 30, 2015 at 20:37

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