# Subspace criteria applied to square matrices, proof help

• karnten07
In summary, there are three subsets of the vector space Mnxn(R) of real (nxn)-matrices: the first subset is the set of square matrices with determinant = 0, the second subset is the set of matrices with rank less than or equal to one, and the third subset is the set of matrices with trace = 0. To show that these subsets are subspaces, it is necessary to prove closure under addition, closure under scalar multiplication, and that the subset contains the 0 vector. However, the first subset is not a subspace because it is not closed under addition. The other two subsets have yet to be proven as subspaces.
karnten07

## Homework Statement

Let n $$\geq$$2. Which of the conditions defining a subspace are satisfied for the following subsets of the vector space Mnxn(R) of real (nxn)-matrices? (Proofs or counterexamples required.

There are three subsets, i will start with the one where The subset V is that of square matrices with determinant = 0.

So in this case i know that these matrices aren't invertible but it isn't a criteria of subspaces for there to be a multiplicative inverse. The first axiom is that there is a zero vector 0v such that x of V time 0v = 0. But how would i write a proof for this?

Then the other axioms i need to show are closure under addition and scalar multiplication, an additive identity element (0), and an additive inverse. I don't need to show multiplicative inverse, no? I mean in any case to show the set is a subspace?

Any help is appreciated, thanks

## The Attempt at a Solution

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I'm not sure where you're getting that first axiom. To be a subspace you need to show the three things you listed in your last paragraph: closure under addition, closure under scalar multiplication, and that the space contains the 0 vector. In this example, the crucial part is closure under addition. Can you prove or give a counterexample to the statement that if det(A) = 0 = det(B) then det(A+B) = 0?

eok20 said:
I'm not sure where you're getting that first axiom. To be a subspace you need to show the three things you listed in your last paragraph: closure under addition, closure under scalar multiplication, and that the space contains the 0 vector. In this example, the crucial part is closure under addition. Can you prove or give a counterexample to the statement that if det(A) = 0 = det(B) then det(A+B) = 0?

Oh, that is great thinking. How about the two matrices,
1 1
0 0

and

1 0
1 0

2 1
1 0

and the inverse is ad - bc = 0 - 1 = -1 so it isn't of the set V.

Thats great , thanks. To show that it has a zero vector:

0v =
00
00

Then A of the set V,

0v.A = 0
0v.a1 a2 = 0
a3 a4
= 0 0
0 0 = 0

Does that satisfy as a proof for the existence of the zero vector?

The 0 vector is characterized by A+0 = A = 0+A For all A in the vector space. It doesn't have to do with multiplication since, in general, you don't have a notion of multiplying vectors (you just happen to in this case).

eok20 said:
The 0 vector is characterized by A+0 = A = 0+A For all A in the vector space. It doesn't have to do with multiplication since, in general, you don't have a notion of multiplying vectors (you just happen to in this case).

Sorry my mistake, so it should be easy to prove that one.

Okay so that is the first subset dealt with. The two are where the square matrices have:

rank(A) $$\leq$$1

and trace(A) = 0

Where trace(A) denotes the sum aij of the diagonal element of the matrix A =(aij)

Im unsure about rank right now so will read on it, but i have no idea where to start with the trace = 0.

Oh, i can use that trace(A+B) = trA +trB
and Tr(xA) = x(TrA)

To show that scalar multiplication is closed for the case when det(A)=0, how do i show a general determinant for any sized square matrix? Because i need to show the determinant is still zero after sclr multiplication.

It is a well known fact that det(AB) = det(A)det(B). Further, if c is a real number and M is a matrix then cM = (cI)M where I is the identity matrix. Using these facts can you deduce what det(cM) is if you know det(M)?

eok20 said:
It is a well known fact that det(AB) = det(A)det(B). Further, if c is a real number and M is a matrix then cM = (cI)M where I is the identity matrix. Using these facts can you deduce what det(cM) is if you know det(M)?

It has to be zero then, so by multiplying the scalar by the identity allows me to calculate the identity for cM using the rules for dterminants of matrices multiplied together. Thats so great, thanks guys.

The only one I am having trouble with now is the rank matrices, where rank is less than or equal to one. What does this mean and where do i start?

If rank is less than or equal to one, doesn this mean that the set is made up of 2x2 matrices with just one element in them that is a number other than zero?

karnten07 said:
Oh, that is great thinking. How about the two matrices,
1 1
0 0

and

1 0
1 0

2 1
1 0

and the inverse is ad - bc = 0 - 1 = -1 so it isn't of the set V.

Thats great , thanks. To show that it has a zero vector:

0v =
00
00

Then A of the set V,

0v.A = 0
0v.a1 a2 = 0
a3 a4
= 0 0
0 0 = 0

Does that satisfy as a proof for the existence of the zero vector?
There is no longer any point in worrying about the zero vector. You have shown, by counterexample, that the set of matrices with determinant 0 is not closed under addition. You don't need to look at the other conditions- if it doesn't satisfy one, it can't satisfy all and so is not a subspace.

By the way, you never have to show, separately, that the 0 vector is in the subspace. After you have shown the set is closed under scalar multiplication, it follows from the fact that $0\vec{v}= \vec{0}$.

## 1. What is the concept of subspace criteria applied to square matrices?

Subspace criteria applied to square matrices is a mathematical concept that involves determining whether a subset of a larger vector space is itself a subspace. In the case of square matrices, this involves checking if the subset of matrices satisfies certain conditions to be considered a subspace of the larger space of all square matrices.

## 2. Why is it important to understand subspace criteria applied to square matrices?

Understanding subspace criteria applied to square matrices is important because it allows us to identify and analyze certain properties and relationships within a set of square matrices. This can be useful in solving mathematical problems, as well as in various applications in fields such as physics, engineering, and computer science.

## 3. What are the conditions that must be satisfied for a subset of square matrices to be considered a subspace?

To be considered a subspace, a subset of square matrices must satisfy three conditions: closure under addition, closure under scalar multiplication, and containing the zero vector. This means that when two matrices from the subset are added or multiplied by a scalar, the resulting matrix must also be in the subset, and the subset must include the zero matrix.

## 4. How can subspace criteria be applied in practical situations?

Subspace criteria can be applied in various practical situations, such as in linear algebra problems, where it can help determine whether a set of vectors or matrices forms a basis for a vector space. It can also be used in physics to analyze the motion of objects in a given space, or in computer science to optimize algorithms by identifying subspaces of data that can be manipulated more efficiently.

## 5. Can subspace criteria be applied to non-square matrices?

Yes, subspace criteria can be applied to any type of matrix, not just square matrices. However, the criteria for a subset to be considered a subspace may differ depending on the type of matrices involved. For example, for rectangular matrices, the number of rows and columns must be fixed, while for diagonal matrices, the diagonal entries must all be zero for the subset to be a subspace.

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