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Which one of these statements is wrong ?

Yankel

Active member
Jan 27, 2012
398
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 ?
 

Yankel

Active member
Jan 27, 2012
398
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 ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
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?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $\mathbb{R}$.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

Yankel

Active member
Jan 27, 2012
398
thank you both !

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

(1+1!=1)

:eek: