How to Show that Linear transformation is invertible?

Question

Let $T_1,T_2$ be two linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$. Let $\{x_1,\cdots , x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1x_i \neq 0$ for every $i=1,2,\cdots, n$ and that $x_i \bot \ker T_2$ for every $i=1,2,\cdots, n$. Which of the following is/are necessarily true?

$T_1$ is invertible.
$T_2$ is invertible.
both $T_1,T_2$ are invertible.
neither $T_1$ nor $T_2$ is invertible.

My Approch:

Given dimension of $\mathbb{R}=n$ and

$x_i\perp Ker(T_2)$

So, $\text{Ker}T_2=0$

Row space and null space are perpendicular to each other.

From Rank Nullity Theorem:

let T : V → W be a linear map. Then the rank of T is the dimension of the image of T and the nullity of T is the dimension of the kernel of T, so we have

$\text{dim}(\text{Im}(T))+ \text{dim}(\text{Ker}(T))=\text{dim}(V)$

Applying above result we get,

$\text{dim}(\text{Im}(T))+ 0=n$

Therefore, $T_2$ is one-one.

I could not solve this problem completely. what to do next?

Is that $m$ actually an $n$? Because for $n \neq m$, invertibility cannot occur. — Crostul, Apr 03 '16 at 08:07
@Crostul The OP does not assume that the target space is the same as the source space. — Noah Schweber, Apr 03 '16 at 08:10
@Crostul This question was asked in the CSIR NET examination and In the answer key, option 2 was marked as the correct choice. — Learner, Apr 03 '16 at 08:18
I can't see how option (2) was marked as true. If it were given $;m=n;$ then yes, as in finite dimension, an operator is injective if and only if it is surjective if and only if it is an isomorphism. The map $;T_2:\Bbb R\to\Bbb R^2;$ defined by $;Tx=(x,0)^t;$ is injective, yet $;T_2;$ can't possibly be invertible as it isn't surjective. — DonAntonio, Apr 03 '16 at 08:35
@Learner, if you meant CSIR NET December 2013 question, no, the key is not wrong. It asks for m=n. The original question can be found here: Is Linear transformations $T_1,T_2:\Bbb R^n\to\Bbb R^n$ Invertible? — Jesse P Francis, Jun 18 '16 at 12:59

score 1 · Accepted Answer · answered Apr 03 '16 at 08:33

Neither of the options are necessarily true.

$T_1$ and $T_2$ can both be invertible, for example if $n=m$ and $T_1$ and $T_2$ are both the identity map.

On the other hand, both $T_1$ and $T_2$ can fail to be invertible, for example if $n=1$, $m=2$ and $T_1(x)=(x,0)$ and $T_2(x)=(x,0)$.

Noah Schweber · Answer 2 · 2016-04-03T08:27:22.303

0

You are correct that $T_2$ is injective (although rank nullity is overkill: just note that $T_2(a)=T_2(b)$ implies $T_2(a-b)=0$, so if $ker(T_2)$ is trivial then $T_2$ is injective). However, it might not be onto (hence not invertible): what if $m>n$?

For $T_1$, make it a bit simpler: what are some linear transformations from $\mathbb{R}^2$ to $\mathbb{R}^1$? For example, let $\{e_1, e_2\}$ be the standard basis for $\mathbb{R}^2$, and let's say we're only looking at linear transformations which send each $e_i$ to either $0$ or $1$ - how many such linear transformations are there? Does one of them in particular seem relevant?

edited Apr 03 '16 at 08:27

answered Apr 03 '16 at 08:10

Noah Schweber

245,398

I have shown that $T_2$ is one-one. How to show that It's onto. – Learner Apr 03 '16 at 08:12
@Learner Whoops, it's late and I mistook "invertible" for "injective". – Noah Schweber Apr 03 '16 at 08:27

How to Show that Linear transformation is invertible?

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