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Consider the linear transformation given by L(x)=Ax where A is the matrix $$\left(\begin{matrix}1&0&3\\2&1&2&\\1&2&-5\\2&1&1\end{matrix}\right)$$

Using the dimension theorem show that L is injective

Dimension Theorem Let V be a vector space over a field F of finite dimension and let L : V → W be a linear transformation from V to another vector space W over F. Then:

Dim(Ker(L)) + Dim(Im(L)) = Dim(V)

I know that L is injective if and only if dim(Ker(L)) = 0; how can I show that dim(Ker(L)) = 0 using the dimension theorem?

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  • $\begingroup$ Prove that the $\text{Dim(Im(L))}=\text{Dim(V)}$. This imply that $\text{Dim(Ker(L))}=0$ $\endgroup$ – Carlos Jiménez May 2 '17 at 2:13
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Verify how many linearly independent rows has the matrix, and substract that to the total number of rows, that's the Im(L) dimension. So, from the matrix you can see Dim(V)=3. Then , if Dim(Ker[L])=3- Dim(Im[L]) = 0 $\Rightarrow$ L is inyective

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The matrix isn’t zero and its columns obviously aren’t multiples of each other, so the rank is at two. Looking at the second and fourth rows, it should be clear that there’s no linear combination of the first two columns that can equal the third, so the rank of the matrix must be three. You can take it from there.

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Convert the given matrix, say $A$ into row Echelon form to compute the rank of $A$ which will come out to be $3$. Now by rank-nullity theorem

$Rank(A) + nullity (A) = 3$, which implies that $nullity (A) = dim (Ker A) = 0$. Note that, since $A$ is a $4\times 3$ matrix, considering it as a linear transformation, $A$ will be a map from $\mathbb{R}^3$ to $\mathbb{R}^4 $.

If you are not aware of Echelon form of matrix, you can ask in the thread.

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