# Same Dim. and Isomorphism

TranscendArcu

## Homework Statement

A theorem in my book states: If V, W are finite dimensional vector spaces that are isomorphic, then V, W have the same dimension. I wrote a proof but it is different from the proof given in my book, and I'd like to know if it's right.

## The Attempt at a Solution

Let $\left\{A_1, ...,A_n\right\}$ be a basis for V. Because V,W are isomorphic, we know $ImT = W$, which implies that $dim(ImT) = dim(W)$. We also know $KerT = \left\{ 0 \right\}$ and hence $dim(KerT) = 0$ We know $dim(V) = n = dim(ImT) + dim(KerT)= dim(ImT) + 0 = dim(ImT)$. Thus, $dim(ImT) = n = dim(W)$

Is that right?

sunjin09

## Homework Statement

A theorem in my book states: If V, W are finite dimensional vector spaces that are isomorphic, then V, W have the same dimension. I wrote a proof but it is different from the proof given in my book, and I'd like to know if it's right.

## The Attempt at a Solution

Let $\left\{A_1, ...,A_n\right\}$ be a basis for V. Because V,W are isomorphic, we know $ImT = W$, which implies that $dim(ImT) = dim(W)$. We also know $KerT = \left\{ 0 \right\}$ and hence $dim(KerT) = 0$ We know $dim(V) = n = dim(ImT) + dim(KerT)= dim(ImT) + 0 = dim(ImT)$. Thus, $dim(ImT) = n = dim(W)$

Is that right?

This is the proof in my book

TranscendArcu
This is the massive proof in my text. It's seems so unintuitive to me.

http://desmond.imageshack.us/Himg696/scaled.php?server=696&filename=screenshot20120202at557.png&res=medium [Broken]
http://desmond.imageshack.us/Himg832/scaled.php?server=832&filename=screenshot20120202at558.png&res=medium [Broken]
http://desmond.imageshack.us/Himg215/scaled.php?server=215&filename=screenshot20120202at558.png&res=medium [Broken]

I mean, seriously.

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sunjin09
This is the massive proof in my text. It's seems so unintuitive to me.

http://desmond.imageshack.us/Himg696/scaled.php?server=696&filename=screenshot20120202at557.png&res=medium [Broken]
http://desmond.imageshack.us/Himg832/scaled.php?server=832&filename=screenshot20120202at558.png&res=medium [Broken]
http://desmond.imageshack.us/Himg215/scaled.php?server=215&filename=screenshot20120202at558.png&res=medium [Broken]

I mean, seriously.

My book is written by a physicist, for non-mathematicians, (explicitly stated in the book title)

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Homework Helper
Gold Member
This is the massive proof in my text. It's seems so unintuitive to me.
What aspect of it do you find unintuitive? It seems straightforward and a natural way to proceed if you don't want to appeal to the rank-nullity theorem (which itself requires a non-trivial proof) as you did in your solution. Maybe the book hadn't proved the latter theorem yet?

Also, your text's proof gives some additional information. Instead of merely verifying that the two spaces have the same dimension, it shows you how to construct a basis for the target space.

TranscendArcu
I'm just working some practice problems here and I'd like to get my work checked. In particular: http://desmond.imageshack.us/Himg638/scaled.php?server=638&filename=screenshot20120202at613.png&res=medium [Broken]

Let $X = (x_1,x_2,x_3), Y = (y_1,y_2,y_3) \in R^3$. Let $a,b \in R$. $T(aX + bY) = T(ax_1 + by_1,ax_2 + by_2,ax_3 + by_3) = (ax_2 + by_2,0,ax_3 + by_3) = (ax_2,0,ax_3) + (by_2,0,by_3) = a(x_2,0,x_3) + b(y_2,0,y_3) = aT(X) + bT(Y)$. Which shows that T is linear.

Vertors in $KerT$ have the form $(x,0,0)$ where $x \in R$. Thus, the standard basis will do for all vectors in the kernel. Namely $\left\{(1,0,0)\right\}$, which has dimension 1.

Vectors in $ImT$ have the form $(y,0,z)$ where $y,z \in R$. Again the standard basis will do for all vectors in kernel. Namely, $\left\{(1,0,0),(0,0,1) \right\}$, which has dimension 2.

Vectors in T(U) have the form $(0,0,z)$ where $z \in R$. This is just the z-axis and we can use $\left\{ (0,0,1) \right\}$ as a basis (which is dimension 1).

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TranscendArcu
What aspect of it do you find unintuitive? It seems straightforward and a natural way to proceed if you don't want to appeal to the rank-nullity theorem (which itself requires a non-trivial proof) as you did in your solution. Maybe the book hadn't proved the latter theorem yet?

Also, your text's proof gives some additional information. Instead of merely verifying that the two spaces have the same dimension, it shows you how to construct a basis for the target space.

Maybe unintuitive was the wrong word. Personally, I'm not a fan of that proof. Partly because it's long, partly because I didn't think of it, and partly because it uses the fundamental definition for a basis (namely, that it spans and is linearly independent). I like my proof more because it is short and relies on the simple equation dimV = dimKerT + dimImT.

Homework Helper
Gold Member
Maybe unintuitive was the wrong word. Personally, I'm not a fan of that proof. Partly because it's long, partly because I didn't think of it, and partly because it uses the fundamental definition for a basis (namely, that it spans and is linearly independent). I like my proof more because it is short and relies on the simple equation dimV = dimKerT + dimImT.

Fair enough, but the equation dimV = dimKerT + dimImT requires a similar, even longer proof, so you have simply hidden the details by using that theorem. Nothing logically wrong with that, as long as "dimV = dimKerT + dimImT" has already been proved at this point.

Homework Helper
Gold Member
I'm just working some practice problems here and I'd like to get my work checked. In particular: http://desmond.imageshack.us/Himg638/scaled.php?server=638&filename=screenshot20120202at613.png&res=medium [Broken]

Let $X = (x_1,x_2,x_3), Y = (y_1,y_2,y_3) \in R^3$. Let $a,b \in R$. $T(aX + bY) = T(ax_1 + by_1,ax_2 + by_2,ax_3 + by_3) = (ax_2 + by_2,0,ax_3 + by_3) = (ax_2,0,ax_3) + (by_2,0,by_3) = a(x_2,0,x_3) + b(y_2,0,y_3) = aT(X) + bT(Y)$. Which shows that T is linear.

Vertors in $KerT$ have the form $(x,0,0)$ where $x \in R$. Thus, the standard basis will do for all vectors in the kernel. Namely $\left\{(1,0,0)\right\}$, which has dimension 1.

Vectors in $ImT$ have the form $(y,0,z)$ where $y,z \in R$. Again the standard basis will do for all vectors in kernel. Namely, $\left\{(1,0,0),(0,0,1) \right\}$, which has dimension 2.

Vectors in T(U) have the form $(0,0,z)$ where $z \in R$. This is just the z-axis and we can use $\left\{ (0,0,1) \right\}$ as a basis (which is dimension 1).

This all looks correct to me.

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