# Thread: Determinants - Proving that |A|*|B|=|A*B|

1. ## Determinants - Proving that |A|*|B|=|A*B|

How can I prove this:

?

BTW - If you have sites dedicated for explanations of mathematical proofs, then I'll be glad to have a link to them. Thank you very much!

2. the usual proof is long and cumbersome. the sophisticated one uses things you do not know. for reading good articles on the web plz see the blog "the unapologetic mathematician". i wish i knew this blog when i was starting maths.

3. Well, I did found one proof, and it is long and sophisticated, and since I have my first math-test this Monday I just wished there would be an 'easier way', which is something that is many-times possible in mathematical proofs.

I need to know the proofs of the lemmas and theorems in Linear Algebra 1, and the rest of them I was able to find and understand in books and google.

BTW - Any tips for a first test? what to do if I don't seem to remember a proof by heart, and too stressful to figure it out at the moment?

Thank you very much!

4. Given it's your first maths test it is VERY doubtful whether you'll be asked to prove this.

I tried to post up a proof of it on ProofWiki:

Determinant of Matrix Product - ProofWiki

but I had to abandon it because of lack of finding a decent way of getting the sign right without a lot of extra complication.

Check out ProofWiki btw, it's growing all the time and may well be what you're looking for.

5. Well... Thank you very much, but where I study is a special program for teenagers in high-school, and they expect us to work very hard... so I prefer knowing as much as I can

BTW - I also need the name of these theorems (or a link to a proof):

If V,W are subspaces:
dim(V+W)=dim(V)+dim(W)-dim(V∩W)

If T is a linear transformation, then:
dim(Im(T))+dim(Ker(T))=dim(V)

(I don't study in English, so I need to figure out their names in order to be able to look up for them)

Thank you

6. Got this one:

Dimension of Sum and Intersection of Vector Spaces - ProofWiki

I don't think we've got the other one up on that site yet - our coverage of linear transformations is a bit limited yet. I think you're probably way ahead of me already anyway! Best of luck.

I also need the name of these theorems (or a link to a proof):

If V,W are subspaces:
dim(V+W)=dim(V)+dim(W)-dim(V∩W)

If T is a linear transformation, then:
dim(Im(T))+dim(Ker(T))=dim(V)
I don't know of a name for the first of those theorems. The second one is often known as the "rank plus nullity theorem".

Well... Thank you very much, but where I study is a special program for teenagers in high-school, and they expect us to work very hard... so I prefer knowing as much as I can

BTW - I also need the name of these theorems (or a link to a proof):

If V,W are subspaces:
dim(V+W)=dim(V)+dim(W)-dim(V∩W)

If T is a linear transformation, then:
dim(Im(T))+dim(Ker(T))=dim(V)

(I don't study in English, so I need to figure out their names in order to be able to look up for them)

Thank you
Let $V , W$ be vector spaces over a field $F$ and $T:V \to W$ a linear transformation. Then, $dim(V) = dim (Im (T)) + dim (Ker (T))$

Proof:

Let $k = dim(ker (T))$ , $n = dim(V)$ and let $(v_1, v_2, ... , v_k)$ be a base of $Ker (T)$. We will complete it to $(v_1, v_2, ... , v_k, v_{k+1}, ... , v_n)$ - a base of V. (we can do this since $v_1,...,v_k$ are linearly independent)

We want to show that $dim(Im (T)) = n-k$. If we prove that $(v_{k+1},...,v_n)$ is a base of $Im (T)$, we will be done. First, we will show that $(v_{k+1},...,v_n)$ span $Im (T)$, and then we will show that they are linearly independent, and thus a base to $Im (T)$.

(I) Let $v \in V$, then $v = a_1v_1 + a_2v_2+...+a_kv_k+a_{k+1}v_{k+1}+...+a_nv_n$ where $a_1,...,a_n$ are scalars from $F$. Then:
$T(v) = T(a_1v_1 + ... + a_kv_k + a_{k+1}v_{k+1} + ... + a_nv_n)$ and thus $T(v) = a_1T(v_1) + a_2T(v_2) + ... + a_kT(v_k) + a_{k+1}T(v_{k+1}) + ... + a_nT(v_n)$, but $v_1,v_2,...,v_k \in Ker(T) \Rightarrow T(v) = a_{k+1}T(v_{k+1}) + ... + a_nT(v_n)$ and this gives us that $a_{k+1},...,a_n$ spans $Im(T)$

(II) We will now show that $a_{k+1},...,a_n$ are linearly independent. Assume $\beta_{k+1}v_{k+1} + ... + \beta_nv_n = 0$. We want to show that $\beta_{k+1} = \beta_{k+2} = ... = \beta_n = 0$

The RHS of this equation is also equal to $T(v_1 + v_2 + ... +v_k) = \gamma_1v_1 + \gamma_2v_2 + ... + \gamma_kv_k$ since $v_1,...,v_k \in Ker(T)$. Substitute these expressions and we get:

$\beta_{k+1}v_{k+1} + ... + \beta_nv_n = \gamma_1v_1 + ... + \gamma_kv_k$ This gives us:

$-\gamma_1v_1 - ... - \gamma_kv_k + \beta_{k+1}v_{k+1} + ... + \beta_nv_n = 0$ but we know $(v_1,...,v_n)$ is a base to V and thus linearly independent, so we get:

$\gamma_1 = \gamma_2 = ... = \gamma_k = \beta_{k+1} = \beta_{k+2} = ... = \beta_n = 0$ and so $(v_{k+1},...,v_n)$ is a base of $Im(T)$ and thus $dim (Im(T)) = dim(V) - dim(Ker(T))$ as required.

9. Ah - looks like we do have this in there after all:

Sum of Nullity and Rank of Linear Transformation - ProofWiki

10. Thank you all, very much!! I really appreciate your good will to help!