# Standard Matrices and basis vectors

• Jun 18th 2010, 04:33 PM
Jukodan
Standard Matrices and basis vectors
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. Attachment 17909
• Jun 18th 2010, 06:47 PM
Ackbeet
Can you find a matrix representation of T? I think that would get you started.
• Jun 18th 2010, 07:55 PM
Jukodan
Quote:

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.
• Jun 18th 2010, 08:00 PM
Ackbeet
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?
• Jun 18th 2010, 08:36 PM
Jukodan
The part I don't understand. Would it be a 2x2 or 3x3? I would assume it comes out to be $\begin{bmatrix}
1\, \, \, 0 & \\
0\, \, \, 1 &
\end{bmatrix}$
• Jun 19th 2010, 05:41 AM
HallsofIvy
Quote:

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 $R^3$ 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.
• Jun 20th 2010, 06:47 PM
Jukodan
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?
• Jun 20th 2010, 07:58 PM
HallsofIvy
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?
• Jun 22nd 2010, 10:44 AM
Jukodan
The standard matrix is ..... $\begin{bmatrix}
3\:-1 & \\
-1\;\; 1& \\
0\;\;5&
\end{bmatrix}$
• Jun 22nd 2010, 11:51 AM
HallsofIvy
Quote:

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

Yes, that is correct- very good!
• Jun 23rd 2010, 09:28 PM
Jukodan
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.