# Lin Alg Proofs and Counterexamples

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• May 2nd 2010, 01:13 PM
dwsmith
Lin Alg Proofs and Counterexamples
I have compiled 57 prove or disprove lin alg questions for my final; however, these may be useful to all.

Contributors to some of the solutions are HallsofIvy, Failure, Tikoloshe, jakncoke, tonio, and Defunkt.

If you discover any errors in one of the solutions, then feel free to reply with the number and correction.
• May 2nd 2010, 03:20 PM
Defunkt
Test 2, Q1:
You are supposed to prove here that $det(AB) = det(BA)$ for any matrices A,B. Why, then, do you assume that B is singular?

Test 2, Q2:
You are meant to give a counter example here, like you did in Q3.

Test 2, Q4:
Your proof is incorrect. You are saying, essentially, that if A,B are row equivalent then $A=E_1E_2...E_k$ where $E_1E_2...E_k=B$ but then $A=E_1E_2...E_k=B \Rightarrow A=B$ and clearly not any two row equivalent matrices are equal. The correct formulation is that $A=E_1E_2...E_kB$ with $E_i$ elementary matrices.
You should also note that this proposition is not true: Take $A=I_2$ and do the row operation $R_1 \leftrightarrow R_2$ to get B, then: $det(A) = 1, det(B) = -1 \neq det(A)$

Test 2, Q5:
This is also incorrect. I don't see how $\mathbb{R}^2$ supposedly not having a third coordinate has anything to do with closure under addition. You should note that in $\mathbb{R}^3$, $\mathbb{R}^2$ would be $\{(x,y,z) \in \mathbb{R}^3 : z = 0\}$. It should now be obvious that $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3$ (and this makes much more sense, does it not?)

Test 4, Q3:
As already mentioned, if you think a proposition is false you need to provide a counter example, not just say that it may not always be true.

Test 4, Q5:
0 is not a positive number.
• May 2nd 2010, 03:26 PM
dwsmith
Quote:

Originally Posted by Defunkt
Test 2, Q1:
You are supposed to prove here that $det(AB) = det(BA)$ for any matrices A,B. Why, then, do you assume that B is singular?

By doing so, we can show that det(AB)=det(BA).

det(A)det(B)=det(A)0=0
det(B)det(A)=0det(A)=0
det(AB)=det(BA)
• May 2nd 2010, 03:32 PM
dwsmith
Quote:

Originally Posted by Defunkt
Test 4, Q3:
As already mentioned, if you think a proposition is false you need to provide a counter example, not just say that it may not always be true.

This answer comes directly from the def of an orthonormal matrices.
• May 2nd 2010, 03:33 PM
Defunkt
But you're still only showing it for a singular B... What if both A and B are non-singular? Your proof doesn't cover that case. det(AB) doesn't always equal 0.

Quote:

This answer comes directly from the def of an orthonormal matrices.
Yes, but I believe the purpose of those questions is giving a counter-example. I don't think you would get a full grade for only explaining why it is not always true.
• May 2nd 2010, 03:38 PM
dwsmith
Quote:

Originally Posted by Defunkt
Yes, but I believe the purpose of those questions is giving a counter-example. I don't think you would get a full grade for only explaining why it is not always true.

Giving a counterexample was not a requirement for this file I created. If it is easier to use a counterexample, I did. If it is easy to use a definition to show it doesn't work, I did that too.

This is just something I created. It doesn't necessarily mean a counterexample is mandatory.
• May 2nd 2010, 03:44 PM
dwsmith
Quote:

Originally Posted by Defunkt
Test 2, Q5:
This is also incorrect. I don't see how $\mathbb{R}^2$ supposedly not having a third coordinate has anything to do with closure under addition. You should note that in $\mathbb{R}^3$, $\mathbb{R}^2$ would be $\{(x,y,z) \in \mathbb{R}^3 : z = 0\}$. It should now be obvious that $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3$ (and this makes much more sense, does it not?)

It probably could be worded better but check out this link.
Why R2 is not a subspace of R3? [Archive] - Physics Forums
• May 2nd 2010, 03:45 PM
dwsmith
Quote:

Originally Posted by Defunkt
Test 4, Q5:
0 is not a positive number.

We discussed this in class and decided to disregard the semantic issue about 0 to instead answer the question.
• May 2nd 2010, 03:47 PM
dwsmith
Quote:

Originally Posted by Defunkt
But you're still only showing it for a singular B... What if both A and B are non-singular? Your proof doesn't cover that case. det(AB) doesn't always equal 0.

If you provide the proof you deem necessary, I will update the file.
• May 2nd 2010, 04:14 PM
dwsmith
Quote:

Originally Posted by Defunkt
Test 2, Q1:
Test 2, Q4:
Your proof is incorrect. You are saying, essentially, that if A,B are row equivalent then $A=E_1E_2...E_k$ where $E_1E_2...E_k=B$ but then $A=E_1E_2...E_k=B \Rightarrow A=B$ and clearly not any two row equivalent matrices are equal. The correct formulation is that $A=E_1E_2...E_kB$ with $E_i$ elementary matrices.
You should also note that this proposition is not true: Take $A=I_2$ and do the row operation $R_1 \leftrightarrow R_2$ to get B, then: $det(A) = 1, det(B) = -1 \neq det(A)$

You are correct here and I will update.
• May 2nd 2010, 04:17 PM
Defunkt
Quote:

Originally Posted by dwsmith
If you provide the proof you deem necessary, I will update the file.

Of course, $det(AB)=det(A)det(B)$ is only correct when A,B are square matrices. Otherwise, this is not true. I'll leave it to you to find a counter example in this case.
• May 2nd 2010, 04:21 PM
dwsmith
The row equivalent question has been change and new file uploaded.

Test 2 Q4
• May 2nd 2010, 04:22 PM
dwsmith
Quote:

Originally Posted by Defunkt
Of course, $det(AB)=det(A)det(B)$ is only correct when A,B are square matrices. Otherwise, this is not true. I'll leave it to you to find a counter example in this case.

This probably important, and not mentioned, assume all matrices are nxn for test 2.
• May 2nd 2010, 04:24 PM
dwsmith
The file has been update again for test 2 with the relevant info.
• May 3rd 2010, 12:43 AM
mr fantastic
Quote:

Originally Posted by dwsmith
The file has been update again for test 2 with the relevant info.

Once a final draft has been established I will stick this thread as a useful resource.
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