Proof for Dim(V) = Dim(Kernel(T)) + Dim(Image(T)): Is this enough to prove it?

Question

Prove that $dim(V) = dim(Ker(T)) + dim(Im(T))$

Let $dim(Ker(T)) = k$, and let the basis be {$v_1,...,v_k$}

If $k = 0$, $Ker(T) =$ {$0$}, which means that every vector in $V$ is in $Image(T)$

Let $dim(V) = n$

If $k = n$, then every vector in v is in the kernel. Therefore $Im(T) =$ {$0$}

If $n > k$, we can add $n-k$ vectors to the basis of the kernel to create a basis for $V$

Basis $=$ {$v_1,...,v_k,v_{k+1},...,v_n$}

The Image of T is {$T(v_1),...,T(v_k),T(v_{k+1}),...,T(v_n)$}

As $v_1,...,v_k$ are in the kernel, $T(v_1),...,T(v_k) = 0$

Therefore the basis for the Image of T is {$T(v_{k+1}),...,T(v_n)$}

$dim(Ker(T)) = k , dim(V) = n$

$dim(Im(T)) = n-k$

$dim(V) = dim(Ker(T)) + dim(Im(T))$

Answer 1

No, it is not enough. You also need to show that the vectors in the basis of the image of T are linearly independent. Otherwise, you can't conclude that the dimension of $Im(T) = n - k$ because if the "basis" vectors are linearly dependent then $Im(T) < n - k$.

Answer 2

The existing answer is right in that you need to show the linear independence of $ \{ T( v_{k+1} ), \dots, T( v_n ) \} $ in order to show that the set is a basis, and not just a set that spans $ \mathrm{Im } \; T $. But I thought it would be better to fully show this argument.

Suppose there exists $ a_{k+1}, \dots , a_n \in \mathbb{R} $ not all zero such that $ a_{k+1} T( v_{k+1} ) + \dots + a_n T( v_n ) = 0$. Using linearity of $T$, this is equivalent to $ T ( a_{k+1} v_{k+1} + \dots + a_n v_n ) = 0$, i.e. $ a_{k+1} v_{k+1} + \dots + a_n v_n \in \, \mathrm{ker} \, T $, thus there exists $ b_1, \dots , b_k \in \mathbb{R} $ not all zero such that $ a_{k+1} v_{k+1} + \dots + a_n v_n = b_1 v_1 + \dots + b_k v_k $. But we assumed that $ v_1, \dots, v_n $ was a basis, therefore linearly independent. Contradiction!

Thus $ a_{k+1} = \dots = a_n = 0$, which proves linear independence of $ \{ T( v_{k+1} ), \dots, T( v_n ) \} $