The minor of a matrix is the determinant of the matrix formed by striking out Row and Column . By definition, the corresponding cofactor can be calculated from this minor using the formula

Set ; the formula becomes

.

Therefore, the cofactor must be equal to the opposite of the minor .

The cofactor of a matrix , by definition, is equal to

,

where is the minor of the matrix - the determinant of the matrix formed when Row and Column of are struck out. Therefore, we first find the minor of the matrix by striking out Row 2 and Column 1 of , as shown in the diagram below:

The minor is therefore equal to

This is equal to the product of the upper-left and lower-right elements minus the product of the upper-right to lower-right elements, which is

Setting and in the definition of the cofactor, the formula becomes

,

so

.

f is a homomorphism because it preserves both vector addition and scalar multiplication.

To show this we need to prove both statements

Proof f preserves vector addition

Let u and v be arbitrary vectors in with the form and

Consider . Applying the definition of f we get

This is the same thing as

Hence, f preserves vector addition because

Proof f preserves scalar multiplication

Let u be an arbitrary vector in with the form and let k be an arbitrary real constant.

Consider

This is the same thing we get if we consider

Hence f preserves scalar multiplication because for all vectors u and scalars k.

For example, if is the space of all vectors in of the form , and is the space of all vectors in the form , then is a linear mapping, but is not a subset of , let alone a subspace of .

Yes, is onto because any vector in the codomain, , is the image of a vector from the domain.

The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Since the matrix in question has six columns, for the rank and the nullity to be equal, they must each be 3.

Yes, the set is linearly independent. There are multiple ways to see this

Way 1) Put the vectors into matrix form,

The matrix is already in reduced echelon form. Notice there are three rows that have a nonzero number in them and we started with 3 vectors. Thus the set is linearly independent.

Way 2) Consider the equation

If when we solve the equation, we get then it is linearly independent. Let's solve the equation and see what we get.

Distribute the scalar constants to get

Thus we get a system of 3 equations

Since the vectors are linearly independent.

Since is a matrix, . Since has three linearly independent columns, it must have a column space (and hence row space) of dimension , causing by the definition of rank. Hence.

Only a square matrix - a matrix with an equal number of rows and columns - has an inverse. Therefore, a matrix with five rows and four columns cannot even have an inverse.

Another way to state this definition is that a square matrix is said to diagonalizable if and only if there exists some invertible matrix and diagonal matrix such that .