Step 1

Given that,

The set of all \(n \times n\) matrices whose entries sum to zero is a subset of \(M_{n,n}\)

A nonempty subset W of vector space V is a subspace if it is closed under addition and scalar multiplication.

That is, if u, v in W then u +v lies in W.

If a is any scalar then au also in W.

Let W is the set of all \(n \times n\) matrices whose entries sum to zero.

As n by n zero matrix whose entries sum to zero.

Thus W is non-empty.

Let A and B are two n by n matrix such that all entire sum add up to zero.

\(a_{ij},\dotsc,i,j=1,2,3,\dotsc n\) denotes the entries in the matrix A

\(b_{ij},\dotsc,i,j=1,2,3,\dotsc n\) denotes the entries in the matrix B

Thus, \(\sum_{i,j=1}^n a_{ij}=0\)

\(\sum_{i,j=1}^n b_{ij}=0\)

Step 2

Consider the sum of entries in A+B

\(\sum_{i,j=1}^n a_{ij}+b_{ij}\)

By using summation property,

\(\sum_{i,j=1}^n a_{ij}+\sum_{i,j=1}^n b_{ij}\)

It gives,

0 +0 =0

Thus, \(\sum_{i,j=1}^n a_{ij}+ b_{ij}=0\)

Therefore,

All entries in A+B has sum zero.

A+B lies in W.

Step 3

Now take any scalar u in real number.

Let A is in W.

\(\sum_{i,j=1}^n a_{ij}=0\)

We have to show that uA is in W.

Take the sum of all entries in uA

\(\sum_{i,j=1}^n ua_{ij}\)

\(u\sum_{i,j=1}^n a_{ij}\)

\(u(0)=0\)

Thus, all entries in uA have sum zero.

Thus, uA lies in W.

Thus W is a subspace of \(M_{n,n}\)

Therefore, the set of all \(n \times n\) matrices whose entries sum to zero is a subspace of \(M_{n,n}\).

Given that,

The set of all \(n \times n\) matrices whose entries sum to zero is a subset of \(M_{n,n}\)

A nonempty subset W of vector space V is a subspace if it is closed under addition and scalar multiplication.

That is, if u, v in W then u +v lies in W.

If a is any scalar then au also in W.

Let W is the set of all \(n \times n\) matrices whose entries sum to zero.

As n by n zero matrix whose entries sum to zero.

Thus W is non-empty.

Let A and B are two n by n matrix such that all entire sum add up to zero.

\(a_{ij},\dotsc,i,j=1,2,3,\dotsc n\) denotes the entries in the matrix A

\(b_{ij},\dotsc,i,j=1,2,3,\dotsc n\) denotes the entries in the matrix B

Thus, \(\sum_{i,j=1}^n a_{ij}=0\)

\(\sum_{i,j=1}^n b_{ij}=0\)

Step 2

Consider the sum of entries in A+B

\(\sum_{i,j=1}^n a_{ij}+b_{ij}\)

By using summation property,

\(\sum_{i,j=1}^n a_{ij}+\sum_{i,j=1}^n b_{ij}\)

It gives,

0 +0 =0

Thus, \(\sum_{i,j=1}^n a_{ij}+ b_{ij}=0\)

Therefore,

All entries in A+B has sum zero.

A+B lies in W.

Step 3

Now take any scalar u in real number.

Let A is in W.

\(\sum_{i,j=1}^n a_{ij}=0\)

We have to show that uA is in W.

Take the sum of all entries in uA

\(\sum_{i,j=1}^n ua_{ij}\)

\(u\sum_{i,j=1}^n a_{ij}\)

\(u(0)=0\)

Thus, all entries in uA have sum zero.

Thus, uA lies in W.

Thus W is a subspace of \(M_{n,n}\)

Therefore, the set of all \(n \times n\) matrices whose entries sum to zero is a subspace of \(M_{n,n}\).