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This is a beautiful problem from ISI MStat PSB 2015 Problem 2. We provide detailed solution with prerequisite mentioned explicitly.

For any matrix consider the following three proper-

ties:

- is real valued for all and is upper triangular.
- for all
- for all

Define the following set of matrices:

*= {A: A is and satisfies (1),(2) and (3) above }*

*(a) Show that * is a vector space for any .

(b) Find the dimension of , when n = 2 and n = 3.

- Upper triangular matrix
- Subspace of a vector space
- Dimension of a vector space

(a) To show that * is* a vector space for any

So, here if we can show that is a subspace of the vector space of real matrices with usual matrix addition and scalar multiplication then we are done!

Let's try to show this ,

Putting for all i,j then satisfies all the properties (1),(2) & (3) .

So,

Shall show that (i) for all , and

(ii) for all for all { }-{0} ,

For (i) Take any

Let , D= and if then

Now we will see whether D satisfies all the three properties (1),(2) and (3)

when and

Hence as A and B are upper triangular matrix , D is also an upper triangular matrix .

So it satisfies property (1)

Again , for all and for all ,

then for all as

Hence it satisfies property (2) .

Now we have for all and for all ,then for all as

Hence it satisfies the properties (3)

For (ii) Take any

take any { }-{0}

Let, and if then

Then , when

Hence as A is an upper triangular matrix , K is also an upper triangular matrix .

So it satisfies property (1)

Again , for all then for all as

Hence it satisfies property (2) .

Now we have for all ,then for all as

Hence it satisfies the properties (3)

So, is closed under vector addition and scalar multiplication.

Therefore , is a subspace of the vector space of real matrices with usual matrix addition and scalar multiplication . Hence we are done !

(b) n=2 ,

then , by property (1) , ---(I) by property (2) and ---(II) by property (3) .

Now solving (I) and (II) we get

Giving , = { } hence

n=3

then , by property (1) , ---(I) by property (2) and ---(II) by property (3) .

Now solving (I) and (II) we get (say) then ,

,

Giving , = {t } , .

Hence ,

This is a beautiful problem from ISI MStat PSB 2015 Problem 2. We provide detailed solution with prerequisite mentioned explicitly.

For any matrix consider the following three proper-

ties:

- is real valued for all and is upper triangular.
- for all
- for all

Define the following set of matrices:

*= {A: A is and satisfies (1),(2) and (3) above }*

*(a) Show that * is a vector space for any .

(b) Find the dimension of , when n = 2 and n = 3.

- Upper triangular matrix
- Subspace of a vector space
- Dimension of a vector space

(a) To show that * is* a vector space for any

So, here if we can show that is a subspace of the vector space of real matrices with usual matrix addition and scalar multiplication then we are done!

Let's try to show this ,

Putting for all i,j then satisfies all the properties (1),(2) & (3) .

So,

Shall show that (i) for all , and

(ii) for all for all { }-{0} ,

For (i) Take any

Let , D= and if then

Now we will see whether D satisfies all the three properties (1),(2) and (3)

when and

Hence as A and B are upper triangular matrix , D is also an upper triangular matrix .

So it satisfies property (1)

Again , for all and for all ,

then for all as

Hence it satisfies property (2) .

Now we have for all and for all ,then for all as

Hence it satisfies the properties (3)

For (ii) Take any

take any { }-{0}

Let, and if then

Then , when

Hence as A is an upper triangular matrix , K is also an upper triangular matrix .

So it satisfies property (1)

Again , for all then for all as

Hence it satisfies property (2) .

Now we have for all ,then for all as

Hence it satisfies the properties (3)

So, is closed under vector addition and scalar multiplication.

Therefore , is a subspace of the vector space of real matrices with usual matrix addition and scalar multiplication . Hence we are done !

(b) n=2 ,

then , by property (1) , ---(I) by property (2) and ---(II) by property (3) .

Now solving (I) and (II) we get

Giving , = { } hence

n=3

then , by property (1) , ---(I) by property (2) and ---(II) by property (3) .

Now solving (I) and (II) we get (say) then ,

,

Giving , = {t } , .

Hence ,

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