All Linear Algebra Resources
Example Questions
Example Question #14 : Eigenvalues And Eigenvectors Of Symmetric Matrices
Example Question #15 : Eigenvalues And Eigenvectors Of Symmetric Matrices
,
where is a real number.
For to have two real eigenvalues, what must be true for ?
or
or
can be any real number.
can be any real number.
Any real value of makes a symmetric matrix with real entries. It holds that any eigenvalues of must be real regardless of the value of .
Example Question #14 : Eigenvalues And Eigenvectors Of Symmetric Matrices
Give the set of eigenvalues of in terms of , if applicable.
The only eigenvalue is 0.
The eigenvalues are 0 and .
The only eigenvalue is .
The eigenvalues are and .
The eigenvalues are 0 and .
The eigenvalues are 0 and .
An eigenvalue of is a zero of the characteristic equation formed from the determinant of , so find this determinant as follows:
Subtracting elementwise:
Set the determinant to 0 and solve for :
The determinant can be found by taking the upper-left-to-lower-right product and subtracting the upper-right-to-lower-left product:
,
so the eigenvalues of this matrix are 0 and .
Example Question #1 : Linear Mapping
That last question dealt with isomorphism. This question is meant to point out the difference between isomorphism and homomorphisms.
A homomorphism is a mapping between vector spaces that
Preserves scalar multiplication and is onto
Preserves vector addition and is injective
onto and 1-to-1
Preserves vector addition and scalar multiplication
Preserves vector addition and scalar multiplication
By definition a homomorphism is a mapping that preserves vector addition and scalar multiplication.
Compare this to the previous problem. An isomorphism is a homomorphism that is also 1-to-1 and onto. Therefore isomorphism is just a special homomorphism. In other words, every isomorphism is a homomorphism, but not all homomorphisms are an isomorphisms.
Example Question #2 : Linear Mapping
Consider the mapping . Can f be an isomorphism?
(Hint: Think about dimension's role in isomorphism)
not enough information
Yes
No
No
No, f, cannot be an isomorphism. This is because and have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.
Example Question #3 : Linear Mapping
Isomorphism is an important concept in linear algebra. To be able to tell if a mapping is isomorphic, it is important to be able to know what an isomorphism is.
Let f be a mapping between vector spaces V and W. Then a mapping f is an isomorphism if it is
Onto (surjective)
1-to-1 (injective)
All answers.
Preserves vector addition
Preserves scalar multiplcation
All answers.
An isomorphism is homomorphism (preserves vector addition and scalar multiplcation) that is bijective (both onto and 1-to-1). Therefore an isomorphism is a mapping that is
1) onto
2) 1-to-1
3) Preserves vector addition
4) Preserves scalar multiplcation
Example Question #4 : Linear Mapping
In the previous question, we said an isomorphism cannot be between vector spaces of different dimension. But are all homomorphisms between vector spaces of the same dimension an isomorphism?
Consider the homomorphism . Is f an isomorphism?
No
Yes
Not enough information
Not enough information
The answer is not enough information. The reason is that it could be an isomorphism because it is between vector spaces of the same dimension, but that doesn't mean it is.
For example:
Consider the zero mapping f(x,y)= (0,0).
This mapping is not onto or 1-to-1 because all elements go to the zero vector. Therefore it is not an isomorphism even though it is a mapping between spaces with the same dimension.
Another example:
Consider the identity mapping f(x,y) = (x,y)
This is an isomorphism. It clearly preserves structure and is both onto and 1-to-1.
Thus f could be an isomorphism (example identity map) or it could NOT be an isomorphism ( Example the zero mapping)
Example Question #5 : Linear Mapping
Let f be a mapping such that
Let f be defined such that
Is f 1-to-1 and onto?
No, it is not 1-to-1 and not onto
Yes
No, it is 1-to-1 and not onto
No, it is not 1-to-1 but it is onto
No, it is not 1-to-1 and not onto
f is not 1-to-1 and it is not onto.
f is not onto because all of is not in the image of f. For example, the vector (1,1) is not in the image of f.
f is not 1-to-1. For example, the vector (1,1) and (1,0) both go to the same vector.
Ie f(1,1) = f(1,0). Therefore f is not 1-to-1.
Example Question #6 : Linear Mapping
Let f be a mapping such that
Let f be defined such that
Is f an isomorphism?
(Hint: Consider the zero vector)
No, it is not a homomorphism
Yes
No, it is not 1-to-1
No, it is not onto
No, it is not a homomorphism
f is 1-to-1 and onto but it is not a homomorphism. Therefore it is not an isomorphism. To see this consider f(0,0) = (0,5)
A homomorphism always takes the zero vector to the zero vector. This particular mapping does not. Thus it does not preserve structure ie not a homomorphism.
Example Question #7 : Linear Mapping
The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.
Consider the mapping . Can f be a homomorphism?
Yes
not enough information
No
Yes
The answer is yes. There is no restriction on dimension for homomorphism like there is for isomorphism. Therefore f could be a homomorphism, but it is not guaranteed.
Certified Tutor