Diagonal Matrix

Any given square Matrix where all the elements are zero except for the elements that are present Diagonally is called a Diagonal Matrix. Let’s assume a square Matrix [Aij]n x m can be called as a Diagonal Matrix if Aij= 0, if and only if i ≠ j. That is the Diagonal Matrix definition. There are many other matrices other than the Diagonal Matrix, such as symmetric Matrix, antisymmetric, Diagonal Matrix, etc. In this section, you will be studying Diagonal Matrix definition, the properties of a Diagonal Matrix, sample solved problems of Diagonal Matrix.

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Properties of Diagonal Matrix

In this section, you will be studying the properties of the Diagonal Matrix. 

Property 1: If addition or multiplication is being applied on Diagonal matrices, then the matrices should be of the same order.

Example:

\[ Let A = \begin{bmatrix}1 & 0 \\0 & 5 \end{bmatrix} \]

\[ Let A = \begin{bmatrix}6 & 0 \\0 & 4 \end{bmatrix} \]

\[ A + B = \begin{bmatrix}1 & 0 \\0 & 5 \end{bmatrix}  + \begin{bmatrix}6 & 0 \\0 & 4 \end{bmatrix} \]

\[ A + B = \begin{bmatrix}1+6 & 0+0 \\0+0 & 5+4 \end{bmatrix} \]

\[ A + B = \begin{bmatrix}7 & 0 \\0 & 9 \end{bmatrix} \]

Hence, this is the Diagonal Matrix.

Property 2: When you transpose a Diagonal Matrix, it is just the same as the original because all the Diagonal numbers are 0. 

\[ Let A = \begin{bmatrix}2 & 0 \\0 & 9 \end{bmatrix} then A^{T} = \begin{bmatrix}2 & 0 \\0 & 9 \end{bmatrix} \]

Property 3: Diagonal Matrices are commutative when multiplication is applied. 

\[ Let A = \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \]

\[ Let A = \begin{bmatrix}5 & 0 \\0 & 1 \end{bmatrix} \]

Then\[ A \times B = \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \times \begin{bmatrix}5 & 0 \\0 & 1 \end{bmatrix} \]

\[ A \times B = \begin{bmatrix}15 + 0 & 0 + 0 \\0 + 0 & 0 + 2 \end{bmatrix} \]

\[ A \times B = \begin{bmatrix}15 & 0 \\0 & 2 \end{bmatrix} \]

\[ B \times A = \begin{bmatrix}5 & 0 \\0 & 1 \end{bmatrix} \times \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \]

\[ B \times A = \begin{bmatrix}15 + 0 & 0 + 0 \\0+0 & 0+2 \end{bmatrix} \]

Hence, A x B = B x A

What is the Block Diagonal Matrix?

A Matrix that can be split into multiple different Blocks is called a Block Matrix. In such matrices, the non-Diagonal numbers are 0. Therefore, Aij = 0, where i ≠ j. Such matrices are called Block-Diagonal matrices. 

Here’s an example of a Block Diagonal Matrix:

\[ D = \begin{bmatrix}x_{11} & 0 & 0 & . & . & 0  \\0 & x_{22} & 0 & . & . & 0 \\0 & 0 & x_{33} & . & . & 0 \\. & . & . & . & . & . \\. & . & . & . & . & . \\ 0 & 0 & 0 & 0 & 0 & x_{nm} \end{bmatrix} \]

The Inverse of a Diagonal Matrix

Let us consider a Diagonal Matrix

\[ D = \begin{pmatrix}x_{11} & 0 & 0 \\0 & x_{22} & 0 \\ 0 & 0 & x_{33}\end{pmatrix} \]

The determinants of the above Matrix are 

\[|D|= X_{11} = \begin{pmatrix} x_{22}& 0\\ 0& x_{33} \end{pmatrix} + 0 \begin{pmatrix} 0& 0\\ 0& x_{33} \end{pmatrix} + \begin{pmatrix} 0& x_{22}\\ 0& 0 \end{pmatrix} \]

\[|D|   = x_{11}x_{22}x_{33}\]

\[ Adj |D|= X_{11} \begin{pmatrix}x_{22}x_{33} & 0 & 0 \\0 & x_{11}x_{33} & 0 \\ 0 & 0 & x_{11}x_{33}\end{pmatrix} \]

\[ D^{-1} = \frac{1}{|D|} adj D \]

\[ = \frac{1}{x_{11}x_{22}x_{33}} \begin{pmatrix}x_{22}x_{33} & 0 & 0 \\0 & x_{11}x_{33} & 0 \\ 0 & 0 & x_{11}x_{33}\end{pmatrix} \]

\[ = \begin{pmatrix}\frac{1}{a_{11}} & 0 & 0 \\0 & \frac{1}{a_{22}} & 0 \\ 0 & 0 & \frac{1}{a_{33}}\end{pmatrix} \]

Anti-Diagonal Matrix

If all the numbers in the Matrix are 0 except for the Diagonal numbers from the upper right corner to the lower-left corner, it is called an anti Diagonal Matrix. It is represented as:

\[ \begin{pmatrix} 0& 0 & x_{13} \\0 &x_{22}  & 0 \\ x_{31} & 0 &0 \end{pmatrix}\]

Sample Questions

Question 1: If A = \[ \begin{bmatrix}14 & 0 \\0 & 12 \end{bmatrix} \]

B = \[ \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \]

then apply addition and find out if there is a Diagonal in the Matrix or not. 

Solution: 

Given,

 \[ A = \begin{bmatrix}14 & 0 \\0 & 12 \end{bmatrix} \]

\[ B = \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \]

Then, 

\[ A + B = \begin{bmatrix}14 & 0 \\0 & 12 \end{bmatrix} + \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \]

\[A + B =  \begin{bmatrix}14+3 & 0+0 \\0+0 & 12+2 \end{bmatrix} \]

\[ A + B = \begin{bmatrix}17 & 0 \\0 & 14 \end{bmatrix} \]

Yes, when addition operation is applied between Matrix A and Matrix B, the resultant is Diagonal in the Matrix. 

Question 2: If \[ A = \begin{bmatrix}1 & 0 \\0 & 2 \end{bmatrix} \]

\[ B =  \begin{bmatrix}3 & 0 \\0 & 4 \end{bmatrix} \]

then apply, multiplication, and find out if there is a Diagonal in the Matrix or not. 

Solution: 

Given,

\[ A =  \begin{bmatrix}1 & 0 \\0 & 2 \end{bmatrix} \] and \[B = \begin{bmatrix}3 & 0 \\0 & 4 \end{bmatrix} \]

Then, 

\[ A \times B = \begin{bmatrix}3+0 & 0+0 \\0+0 & 0+8 \end{bmatrix} \]

Therefore, \[ A \times B = \begin{bmatrix}4 & 0 \\ 0 & 8  \end{bmatrix} \]

Yes, when multiplication is applied between Matrix A and Matrix B, the resultant is a Diagonal Matrix. 

Question 3: If \[ A =\begin{bmatrix}2 & 0 \\ 0 & 4  \end{bmatrix} \]

and \[ B = \begin{bmatrix}3 & 0 \\ 0 & 6  \end{bmatrix} \]

show that multiplication is cumulative in Diagonal matrices. 

Solution: 

Given,

\[A = \begin{bmatrix}2 & 0 \\ 0 & 4  \end{bmatrix}\] and \[B = \begin{bmatrix}3 & 0 \\ 0 & 6  \end{bmatrix} \]

Then, 

\[ A \times B = \begin{bmatrix}6+0 & 0+0\\0+0 & 0+24 \end{bmatrix} \]

\[ A \times B = \begin{bmatrix}6 & 0\\0 & 24 \end{bmatrix} \]

\[ B \times A = \begin{bmatrix}6+0 & 0+0\\0+0 & 0+24 \end{bmatrix} \]

\[ B\times A = \begin{bmatrix}6 & 0\\0 & 24 \end{bmatrix} \]

Therefore, \[ A \times B = \begin{bmatrix}6 & 0\\0 & 24 \end{bmatrix} = B\times A = \begin{bmatrix}6 & 0\\0 & 24 \end{bmatrix} \]

Hence Proved.

Yes, multiplication operation is cumulative between Diagonal Matrix A and Diagonal Matrix B.

Question 4: Apply property 2 of a Diagonal Matrix

\[ \begin{bmatrix}5 & 0 \\0 & 10 \end{bmatrix} \]

and show that the transpose of a Matrix is the same as the original. 

Solution:

Given,

\[A = \begin{bmatrix}5 & 0 \\0 & 10 \end{bmatrix} \]

\[ A^{T} = \begin{bmatrix}5 & 0 \\0 & 10 \end{bmatrix} \] (the diagonal number’s sign changes to negative , but -0 = 0 )

Hence Prove that \[A = \begin{bmatrix}2 & 0 \\0 & 9 \end{bmatrix} = A^{T} = \begin{bmatrix}2 & 0 \\0 & 9 \end{bmatrix} \]

Diagonal Matrix

The above-mentioned study material must have given you clarity about the Diagonal Matrix. Students shall know that nothing comes without practice and hence, you're required to attempt a lot of questions so that you're able to test your knowledge and also understand the pattern of questions. 

Students might like to study in different ways and that is understandable. However, for subjects like Mathematics it is best to keep practising and exploring different ways to solve a problem. 

Success comes with hard work, perseverance and consistency. Those who decide to invest their time and effort into going after their goals reach heights with ease. 

Let's get started and understand some of the tricks that can be used to prepare for a tough topic, chapter or even a subject. 

•  Understand the Chapter

You can understand the chapter in a lot of ways. Since students understand in different ways, some by doing, some by teaching, some by being taught and so on, it means that you are required to observe and know what are the best ways from which you can master any particular topic. For practical subjects like Mathematics, it is advisable that you rely most on practice. But this in no way means that you have to overburden yourself by taking up too many reference books or questions, the key is to test your understanding without overdoing anything. 

• Note Down All Formulae

Noting down all the formulas together gives you an overview of all the chapters and that too very quickly. Since it is handy, you may not need to carry these big heavy books everywhere and whenever you need to revise the formulas, you can simply refer to these partners—notes. Noting down all the formulas together also helps to memorize them quickly and that too easily. You may try this technique while preparing for your exams next and this will surely help you in surprising ways! 

• Start by Solving Examples

After you finish with the understanding of the chapter, students always tend to solve hard questions which may get you wrong answers and demotivate you. It further discourages you to take action and try more. You shall always start by practising the solved questions so that you get to understand things by doing them. This would get you correct answers which will surely give your confidence a boost. Students shall understand that jumping onto tougher tasks from the beginning won't get you extra marks but this can surely push your morale down and it would cost you more than you could even think of. And hence, you shall be doing it stepwise.

• Practice as Much as You Can 

You must have heard that practice makes a man perfect. But have you ever applied it? Of course, we are unaware of the powers that practice holds! You should always practice things to make yourself better and better. Maths is a subject of practice, this subject needs the most attention. Since a lot of concepts of mathematics are based on hits and trials, you can only learn after doing. While trying different methods, it is not important that you always get the right answer, you might get wrong answers too but this should not demotivate you. Students shall be trying to keep their spirits high at all times so that they don't lag. 

• Clear All Your Doubts

Clearing all your doubts is as important as practising. Maths is a subject in which you might often feel stuck and that is common. But, the key is to get them cleared as soon as you find them. Letting them build over time would make you weaker at those topics and the sooner you get them cleared, you would be able to crack the hard nuts. Reaching out to a teacher, friend, parent, or even online platforms can be helpful.