Standard Matrices and basis vectors

**Jukodan** · June 18th 2010, 03:33 PM

I am taking an online course for linear algebra and getting a response from my teacher takes a while. Could you guys please help me understand parts a b , and d? I'm not really looking for the answer so much as the steps behind it. Thank you in advance.

Standard Matrices and basis vectors-6.jpg

**Ackbeet** · June 18th 2010, 05:47 PM

Can you find a matrix representation of T? I think that would get you started.

**Jukodan** · June 18th 2010, 06:55 PM

Originally Posted by Ackbeet

Can you find a matrix representation of T? I think that would get you started.

3 -1
-1 1
0 5

I think...but what about e sub1 and e sub 2 that's what's really tripping me up.

**Ackbeet** · June 18th 2010, 07:00 PM

I agree with your representation of $T$ . Now, $e_{i}$ is a standard notation for basis vectors. I would guess that

$e_{1}=\left[\begin{matrix}1\\ 0\end{matrix}\right]$ , and

$e_{2}=\left[\begin{matrix}0\\ 1\end{matrix}\right]$ .

So, what do you suppose $T(e_{1})$ is?

**Jukodan** · June 18th 2010, 07:36 PM

The part I don't understand. Would it be a 2x2 or 3x3? I would assume it comes out to be $\begin{bmatrix} 1\, \, \, 0 & \\ 0\, \, \, 1 & \end{bmatrix}$

**HallsofIvy** · June 19th 2010, 04:41 AM

Originally Posted by Jukodan

The part I don't understand. Would it be a 2x2 or 3x3? I would assume it comes out to be $\begin{bmatrix} 1\, \, \, 0 & \\ 0\, \, \, 1 & \end{bmatrix}$

Neither. And, no, the matrix form is not $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ .

You are told that the linear transfromation is "from $R^2$ to $R^3$ " so you will be multiplying a matrix by a "column matrix" with 2 entries (actually 1 column, 2 rows as a matrix) so there must be two columns in you matrix to be able to multiply. The result is in [itex]R^3[/itex] so it will be a "column matrix" with 3 entries (1 column, 3 rows) which means your matrix must have 3 rows to give 3 results. The matrix representing this linear transformation must be "2 by 3" (2 columns, 3 rows).

Do as Ackbeet suggested: since you titled this "Standard matrices and basis vectors" he, and I, assume that $e_1= \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $e_2= \begin{bmatrix}0 \\ 1\end{bmatrix}$ .

You are told that $T(x_1, y_1)= (3x_1- x_2, -x_1+ x_2, 5x_2)$ so what is T(1, 0)? What is T(0, 1)?

Now use the fact that $\begin{bmatrix}a & b\\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ c \\ e\end{bmatrix}$ and $\begin{bmatrix}a & b\\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}= \begin{bmatrix}b \\ d \\ f\end{bmatrix}$ to find a, b, c, d, e, and f.

**Jukodan** · June 20th 2010, 05:47 PM

So then $T(e_1)=\begin{bmatrix} 3\\ -1\\ 0 \end{bmatrix}$ and $T(e_2)=\begin{bmatrix} -1\\ 1\\ 5 \end{bmatrix}$

Am I following you correctly?

**HallsofIvy** · June 20th 2010, 06:58 PM

Yes, that is correct. And, so now what are a, b, c, d, e, f? In other words, what is the matrix representing this linear transformation in the standard basis?

**Jukodan** · June 22nd 2010, 09:44 AM

The standard matrix is ..... $\begin{bmatrix} 3\:-1 & \\ -1\;\; 1& \\ 0\;\;5& \end{bmatrix}$

**HallsofIvy** · June 22nd 2010, 10:51 AM

Originally Posted by Jukodan

The standard matrix is ..... $\begin{bmatrix} 3\:-1 & \\ -1\;\; 1& \\ 0\;\;5& \end{bmatrix}$

Yes, that is correct- very good!

**Jukodan** · June 23rd 2010, 08:28 PM

I thank both of you very much for guiding me and shedding light on the problem. I understand this a lot better now. Both of you were very helpful, thank you again.

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