Determinants

**adam63** · August 28th 2009, 09:10 AM

How can I prove this:

?

Thank you in advance.

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!

**nirax** · August 28th 2009, 09:18 AM

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.

**adam63** · August 28th 2009, 01:25 PM

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!

**Matt Westwood** · August 28th 2009, 01:57 PM

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.

**adam63** · August 28th 2009, 02:42 PM

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

**Matt Westwood** · August 28th 2009, 10:38 PM

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.

**Opalg** · August 28th 2009, 11:36 PM

Originally Posted by adam63

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".

**Defunkt** · August 29th 2009, 04:54 AM

Originally Posted by adam63

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.

**Matt Westwood** · August 29th 2009, 06:07 AM

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

Sum of Nullity and Rank of Linear Transformation - ProofWiki

**adam63** · August 29th 2009, 06:22 AM

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

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