A few statements about linear maps and vector spaces

Question

I feel bad for posting yet another thread without being able to contribute anything, but these statements have been brought up in my linear algebra lecture without any proof whatsoever for them and it's not clear to me at all how they're the case. If anybody could give me some brief explanation or point me to where I can look at a proof I would be very thankful!

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Let $V,W$ be vector spaces and let $f:V\rightarrow W$ be linear. Then:

1. If $A\subset V$ is a generator for $V$, then $f(A)$ is a generator for $f(V)$.

2. If $\{a_1,...,a_n\}\subseteq V$ is linearly dependent, then $\{f(a_1),...,f(a_n)\}\subseteq W$ is as well.

3. In the general case $\dim(f(V))\leq \dim(V)$. Can you find examples for $=$ and $<$ respectively?

Answer 1

If $A\subset V$ is a generator for $V$, then $f(A)$ is a generator for $f(V)$.

Suppose a set of vectors $A = \{\mathbf{v}_k \}_{k=1}^n$ generates $V$. This means we can write any $\mathbf{v} \in V$ as $\mathbf{v} = \displaystyle \sum_{k=1}^n c_k\mathbf{v}_k$, where the $c_k$'s are constants in some field, such as $\mathbb{R}$. So what about vectors in the space $f(V)$? Vectors in $V$ can be written as the sum above, hence vectors here are of the form $\displaystyle f \left( \sum_{k=1}^n c_k\mathbf{v}_k \right)$. Since $f$ is a linear map, we can rewrite this as:

$$f \left( \sum_{k=1}^n c_k\mathbf{v}_k \right) \ = \ \sum_{k=1}^n f(c_k \mathbf{v}_k) \ = \ \sum_{k=1}^n c_k f(\mathbf{v}_k)$$

This shows that vectors in $f(V)$ can be written as a linear combination of the vectors in the set $\{f(\mathbf{v}_k) \}_{k=1}^n = f(A)$. In other words, $f(A)$ generates $f(V)$.

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If $\{a_1,...,a_n\}\subseteq V$ is linearly dependent, then $\{f(a_1),...,f(a_n)\}\subseteq W$ is as well.

A set of vectors $\{\mathbf{v}_k\}_{k=1}^n$ is called linearly dependent when the equation $\displaystyle \sum_{k=1}^n c_k\mathbf{v}_k = 0$ has a solution where $c_k \neq 0$ for at least one $k$. In other words, $\{\mathbf{v}_k\}_{k=1}^n$ is a linearly dependent set when we can find some nontrivial linear combination of those vectors giving zero. Applying the same trick as above:

$$f \left( \sum_{k=1}^n c_k\mathbf{v}_k \right) = \ \sum_{k=1}^n f(c_k \mathbf{v}_k) \ = \ \sum_{k=1}^n c_k f(\mathbf{v}_k)$$

Because $f$ is linear, $f(\mathbf{0}) = \mathbf{0}$. So a nontrivial linear combination of the $\mathbf{v}_k$'s that gives zero can be "factored through" $f$ as above to get a nontrivial linear combination of $f( \mathbf{v}_k)$'s also giving zero. This shows linear dependence of the set $\big\{ f( \mathbf{v}_k) \big\}_{k=1}^n$.

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In the general case $\dim(f(V))\leq \dim(V)$. Can you find examples for $=$ and $<$ respectively?

This fact is implied by the first fact above. The dimension of a space is merely the minimum number of linearly independent vectors needed to generate the space$^\dagger$. So if $A$ is a linearly independent generating set for $V$, then we have shown above that $f(A)$ must generate $f(V)$. But need $f(A)$ be linearly independent as well? If not: $\text{dim}(f(V)) < \text{dim}(V)$ since we can arrive at a linearly independent generating set for $f(V)$ by removing "superfluous" vectors from $f(A)$.

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$^\dagger$Any two linearly independent generating sets for a vector space will have the same cardinality. This is the dimension theorem for vector spaces.

Answer 2

Another way to answer the second part: you can try proving the contrapositive. If $\{f(a_i)\}$ are independent, then so are $\{a_i\}$. Let

$$c^1 a_1 + \ldots+c^n a_n = 0$$

If $f$ acts on both sides,

$$f(c^1a_1+\ldots+c^na_n) = c^1f(a_1)+\ldots+c^nf(a_n)= f(0) = 0$$

Since $\{f(a_i)\}$ are linearly independent, $c^i=0$ for all $i$.

Third part: Imagine two vector spaces $V$ and $W$ such that $\dim(W) = m \geq n = \dim(V)$ and construct a linear function that maps the basis of $V$ to a set of $n$ linearly independent vectors in $W$. Also try proving that $\dim(f(V)) = \dim(V)$ in this case.

Similarly you can repeat this for a linear function that maps the basis of $V$ to a set of $k$ linearly independent vectors in $W$, where $k<n$.

Answer 3

$3)$ is a consequence of the rank nullity theorem ... since the rank is the dimension of the image, and the nullity is nonnegative; and they add up to $\dim V$...