1
$\begingroup$

Let $S=\{v_1, v_2, v_3\}$ be a basis for $\mathbb{R^3}$ where

$v_1 = (1, 0, 1)$

$v_2 = (-1, 1, -1)$

$v_3 = (-1, 1, 0)$

Let $L: \mathbb{R^3} \mapsto \mathbb{R^2}$ be the linear transformation defined by

$L(v_1) = (2, 7)$

$L(v_2) = (4, 2)$

$L(v_3) = (5, 4)$

1.) Find $L(e_1), L(e_2)$ and $L(e_3)$

2.) Find the standard matrix of $L$

3.) Find a nontrivial vector $u$ in the kernel of $L$. Express your answer in the standard basis.

4.) Write your vector $u$ with respect to the basis $S$.

I'm trying to figure these out, but I don't really understand the concept of getting this information from the basis.

$\endgroup$
  • $\begingroup$ Try expressing $e_1, e_2,e_3$ as a linear combination of $v_1,v_2,v_3$ (trial and error works for small numbers). You are able to get this much, right? $\endgroup$ – астон вілла олоф мэллбэрг Mar 23 '17 at 1:03
  • $\begingroup$ I'm not sure do you mean e1= (1, 0, 0) e2 = (0, 1, 0) e3 = (0, 0, 1) and express these with v1, v2, v3? $\endgroup$ – JanoyCresva Mar 23 '17 at 1:06
  • $\begingroup$ That's right. For example, you can check that $e_2 = v_1+v_2$. Can you find similar expressions for $e_1,e_3$? $\endgroup$ – астон вілла олоф мэллбэрг Mar 23 '17 at 1:09
  • $\begingroup$ e3= -v2+v3 and not sure about e3 $\endgroup$ – JanoyCresva Mar 23 '17 at 1:13
  • $\begingroup$ You can check that $e_1 = v_1 - v_3 + v_2$. Now, by linearity, you will get that:$$ L(e_1) = L(v_1-v_3+v_2) = L(v_1) - L(v_3) + L(v_2)$$. Now, you can substitute and find out the rest of the quantities too. $\endgroup$ – астон вілла олоф мэллбэрг Mar 23 '17 at 1:19
0
$\begingroup$

You must first determine what linear combination of the three basis vectors (v1, v2, and v3) is needed to get each of the standard basis vectors e1, e2, and e3. This is accomplished by solving three separate 3x3 linear systems:

(1) a1v1 + a2v2 + a3v3 = e1

(2) b1v1 + b2v2 + b3v3 = e2

(3) c1v1 + c2v2 + c3v3 = e3

where a=(a1, a2, a3) and b=(b1, b2, b3), and c= (c1, c2, c3) are the scalar coefficients you must determine.

The systems solve easily enough to:

a = (1, 1, -1) b = (1, 1, 0) and c = (0, 1, -1)

which means that:

e1 = v1 + v2 - v3

e2 = v1 - v2

e3 = v2 - v3

Now it is straightforward to compute L(e1) = a1(2,7) + a2(4,2) + a3(5,4) = (1)(2,7) + (1)(4,2) + (-1)(5,4) = (1, 5)

and L(e2) = b1(2,7) + b2(4,2) + b3(5,4) = (6,9)

and L(e3) = c1(2,7) + c2(4,2) + c3(5,4) = (-1,-2)

So the standard matrix of L is then [ (1,5), (6,9), (-1,-2) ]

To find a vector u in the kernel of L you must solve for a non-trivial solution to L(u) = 0

Finally you can write u with respect to the basis S, by using the [a,b,c] transformation you solved earlier:

u(S) = (a,b,c)*u

where u(S) is the u vector expressed with respect to the S basis.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.