Proof for a basis of a linear transformation

Printable View

February 2nd 2010, 10:59 AM
Runty

Proof for a basis of a linear transformation

Suppose that $T,S:R^n \rightarrow R^n$ are inverses.

If { $v_1 ,v_2 ,..., v_k$ } is a basis for a subspace $V$ of $R^n$ and $w_1 = T(v_1), w_2 = T(v_2),..., w_k = T(v_k)$ , prove that { $w_1, w_2,..., w_k$ } is a basis for $T(V)$ .

In addition, give an example to show that this need not be true if T does not have an inverse.
February 2nd 2010, 03:37 PM
Defunkt

Quote:

Originally Posted by Runty

Suppose that $T,S:R^n \rightarrow R^n$ are inverses.

If { $v_1 ,v_2 ,..., v_k$ } is a basis for a subspace $V$ of $R^n$ and $w_1 = T(v_1), w_2 = T(v_2),..., w_k = T(v_k)$ , prove that { $w_1, w_2,..., w_k$ } is a basis for $T(V)$ .

In addition, give an example to show that this need not be true if T does not have an inverse.

First, show that $\{w_1,...,w_k\}$ spans $T(V)$ . Hint:
we can write any $v \in V$ as $v = \sum_{i=1}^k a_iv_i, ~ a_i \in \mathbb{R}$ , therefore: $T(v) = T(\sum_{i=1}^k a_iv_i) = ...$

Now, show that $\{w_1,...,w_k\}$ is linearly independent: Assume it is not, and reach a contradiction.

Therefore it is a basis.
February 4th 2010, 05:30 AM
Runty

Quote:

Originally Posted by Defunkt

First, show that $\{w_1,...,w_k\}$ spans $T(V)$ . Hint:
we can write any $v \in V$ as $v = \sum_{i=1}^k a_iv_i, ~ a_i \in \mathbb{R}$ , therefore: $T(v) = T(\sum_{i=1}^k a_iv_i) = ...$

Now, show that $\{w_1,...,w_k\}$ is linearly independent: Assume it is not, and reach a contradiction.

Therefore it is a basis.

I suppose this answer could work, but I'd like to, if possible, avoid using summation notation.
February 4th 2010, 02:04 PM
Defunkt

Quote:

Originally Posted by Runty

I suppose this answer could work, but I'd like to, if possible, avoid using summation notation.

$v = a_1v_1 + ... + a_kv_k$

$T(v) = T(a_1v_1 + ... + a_kv_k) = a_1T(v_1) + ... + a_kT(v_k)$
February 6th 2010, 06:38 PM
wopashui

Quote:

Originally Posted by Defunkt

$v = a_1v_1 + ... + a_kv_k$

$T(v) = T(a_1v_1 + ... + a_kv_k) = a_1T(v_1) + ... + a_kT(v_k)$

Since $T,S : R^n --> R^n$ which implies it's an nxn matrix, you can just apply the big theorem, to prove it
February 7th 2010, 10:31 AM
Runty

Quote:

Originally Posted by Defunkt

$v = a_1v_1 + ... + a_kv_k$

$T(v) = T(a_1v_1 + ... + a_kv_k) = a_1T(v_1) + ... + a_kT(v_k)$

Okay, that solves the first part. But I still need an example to show that this isn't necessarily true, provided that $T$ does not have an inverse.

Honestly, I find this whole question to be pretty obscure.
February 7th 2010, 11:14 AM
Defunkt

Quote:

Originally Posted by Runty

Okay, that solves the first part. But I still need an example to show that this isn't necessarily true, provided that $T$ does not have an inverse.

Honestly, I find this whole question to be pretty obscure.

Take any transformation that is not invertible; for example, $T:\mathbb{R}^2 \to \mathbb{R}^2$ defined by $T((x,y)) = (x,0)$ . Since T is not invertible, $Ker T \neq \{0\}$ . In fact, $Ker T = \{(x,y) \in \mathbb{R}^2 : x = 0\}$ .

Then, $Ker T$ is spanned by $(0, 1)$ , however $T(0,1) = (0,0)$ which is not a base.

This will work for any transformation that is not invertible:
Since it is not invertible, $Ker T \neq \{0\}$ but for any $w \in Ker T$ , $Tw = 0$ , and therefore the image of any base of $Ker T$ will be mapped to the zero vector.
February 7th 2010, 11:16 AM
Runty

Quote:

Originally Posted by Defunkt

Take any transformation that is not invertible; for example, $T:\mathbb{R}^2 \to \mathbb{R}^2$ defined by $T((x,y)) = (x,0)$ . Since T is not invertible, $Ker T \neq \{0\}$ . In fact, $Ker T = \{(x,y) \in \mathbb{R}^2 : x = 0\}$ .

Then, $Ker T$ is spanned by $(0, 1)$ , however $T(0,1) = (0,0)$ which is not a base.

This will work for any transformation that is not invertible:
Since it is not invertible, $Ker T \neq \{0\}$ but for any $w \in Ker T$ , $Tw = 0$ , and therefore the image of any base of $Ker T$ will be mapped to the zero vector.

By $KerT$ , do you mean determinant? I've never seen the term $Ker$ used before.
February 7th 2010, 08:27 PM
math2009

$ c_1\vec{w}_1+\cdots +c_k\vec{w}_k=0\rightarrow c_1T(\vec{v}_1)+\cdots +c_kT(\vec{v}_k)=0 $

$ \rightarrow c_1T(\vec{v}_1)+\cdots +c_kT(\vec{v}_k)=0 \rightarrow T(c_1\vec{v}_1)+\cdots +T(c_k\vec{v}_k)=0 \rightarrow T(c_1\vec{v}_1+\cdots +c_k\vec{v}_k)=0 $

$ =A(c_1\vec{v}_1+\cdots +c_k\vec{v}_k)=0 \rightarrow c_1\vec{v}_1+\cdots +c_k\vec{v}_k=A^{-1}0=0 $

$ \rightarrow c_1=\cdots =c_k=0 \rightarrow \{\vec{w}_1,\cdots,\vec{w}_k\} $ is linearly independent

$ dim \{\vec{w}_1,\cdots,\vec{w}_k\} = k = dim(V)~ \therefore $ it's basis of V
February 8th 2010, 05:00 AM
HallsofIvy

Quote:

Originally Posted by Runty

By $KerT$ , do you mean determinant? I've never seen the term $Ker$ used before.

"ker(T)" is the "kernel" of T, also called the "null space" of T. It is the subspace of all vectors, v, such that Tv= 0.

If T is invertible, then Tv= 0 gives $T^{-1}T(v)= T^{-1}(0)$ or v= 0. That is, if T is invertible, its kernel (null space) consists only of the 0 vector.

In fact, you can also prove the other way: if the 0 vector is the only vector in the kernel of T, T is invertible.

"null space" is used exclusively in linear algebra. "kernel" of an operator is also used in group theory, ring theory, etc.