MHB Which one of these statements is wrong ?

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The discussion revolves around identifying the incorrect statement among five options related to matrix theory and vector spaces. The participants conclude that statement 2, which claims that an nXn matrix with different numbers on its main diagonal is diagonalized, is false, as it can be demonstrated with a specific matrix example. Statement 5 is confirmed as correct because any linear combination of vectors in a vector space remains within that space. The initial confusion stemmed from a misunderstanding of vector space properties, particularly regarding closure under addition. Ultimately, the participants clarify that the set provided in the example does not constitute a vector space.
Yankel
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Hello

I have a question in which I need to choose the wrong statement. I have 5 statements, and I managed to rule out 3 options so I am left with two.

the options are:

1. The dimension of the 3X3 anti-symmetric matrices subspace is 3.

2. An nXn matrix which has different numbers on it's main diagonal, is diagonalized.

3. A non invertible matrix has an eigenvalue of 0

4. Every matrix has a unique canonical form matrix

5. v1 and v2 are vectors from a vector space V. Then v1-2v2 also belongs to V.

I managed to rule out 1, 3 and 4 (they are correct in my opinion). I don't know which one is not, is it 2 or 5 ?
 
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I think I can rule out 5 too, but that doesn't help me understand why 2 is correct.

am I right to say:

let's assume that V is the space of all vectors of form (1,a,b), then:

v1 = (1,a,b) v2 = (1,c,d)

v1-2v2 = (-1,2-ac,b-2d) which doesn't belong to V ?
 
Hi Yankel!

If 2 vectors belong to a vector space, than any linear combination of those vectors also belongs to that vector space by definition.
Since $v_1-2v_2$ is a linear combination, (5) is correct.

As for (2), which values does a diagonal matrix have that are not on its main diagonal?
 
Yankel said:
2. An nXn matrix which has different numbers on it's main diagonal, is diagonalized.

This is false. Choose for example $A=\begin{bmatrix}{1}&{-2}\\{1}&{-1}\end{bmatrix}\in \mathbb{R}^{2\times 2}$. Its eigenvalues are $\lambda=\pm i\not\in \mathbb{R}$, so $A$ is not diagonalizable on $\mathbb{R}$.
 
Yankel said:
I think I can rule out 5 too, but that doesn't help me understand why 2 is correct.

am I right to say:

let's assume that V is the space of all vectors of form (1,a,b), then:

v1 = (1,a,b) v2 = (1,c,d)

v1-2v2 = (-1,2-ac,b-2d) which doesn't belong to V ?

As ILikeSerena told you, this is true. Your mistake is that the set $\{(1,a,b):a,b\in\mathbb{R}\}$ is not a vector space.
 
thank you both !

Yes, silly example, my set wasn't a vector space since it's not close on addition

(1+1!=1)

:eek:
 
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