Linear algebra

Question

Via independent research, and in your own words, explain what is known as similar matrices.

shaquib1325 · Answer

Similar matrices are explained in 3 different ways:1.Lets say A and B are 2 matrices of size nxn. Then if there exists a non singular S(invertible) matrix of size n such that 
Then A & B are said to be similar matrices.
2.In other words, the matrices which represent same linear map under 2 different bases.
We can find similar matrix of a given matrix using other matrix by the equation shown above.3.If two matrices have same distinct eigen values then they are similar.If a matrix A has k independent eigenvectors, then it is similar to a matrix B which has k Jordan blocks on its diagonal

devalmishra666 · Answer

SIMILAR MATRICES  We shall now introduce the idea of similar matrices and study the properties of similar matrices. DEFINITION  An nxn matrix A is said to be similar to a nxn matrix B if there exists a nonsingular nxn matrix P such that,  P-1 A P = B We then write,  A ∼ B Properties of Similar Matrices (1) Since I-1 A I = A it follows that A ∼ A (2) A ∼ B ⇒ ∃ P, nonsingular such that., P-1 A P = B  ⇒ A = P B P-1 ⇒ A = Q-1 B P, where Q = P-1 is nonsingular  ⇒ ∃ nonsingular Q show that Q-1 B Q = A  ⇒ B ∼ A Thus  A ∼ B ⇒ B ∼ A (3) Similarly, we can show that  A ∼ B, B ∼ C ⇒ A ∼ C. (4) Properties (1), (2) and (3) above show that similarity is an equivalence relation on the set of all nxn matrices. (5) Let A and B be similar matrices. Then there exists a nonsingular matrix P such that  A = P-1 B P.
Now, let CA(λ) and CB (λ) be the characteristic polynomials of A and B respectively. We have,  CA ( ) I A I P BP −1 λ = λ − = λ − P P P BP −1 −1 = λ − = P ( ) I − B P − λ 1 = P I − B P − λ 1 = − λI B since 1 P P 1 − = = CB ( λ ) Thus “ SIMILAR MATRICES HAVE THE SAME CHARACTERISTIC POLYNOMIALS ”. (6) Let A and B be similar matrices. Then there exists a nonsingular matrix P such that  A = P-1 B P Now for any positive integer k, we have  ( )( ) ( ) 14 2 444 4 3 444 k timesk 1 1 1 A P BP P BP ..... P BP − − − = = P-1 Bk P (since PP-1=I) Therefore,  Ak = On ⇔ P-1 Bk P = On ⇔ Bk = On“ Thus if A and B are similar matrices then Ak = On ⇔ Bk = On ”. 7) Let A and B be any two square matrices of the same order, and let  p(λ) = a0 + a1λ + &hellip;.. + akλ be any polynomial. Then ( ) k p A = a0 I + a1A + .....+ ak A a I a P BP a P B P a P B Pkk1 2 1210 1 ..... − − − = + + + + P [a I a B a B a B ]Pk = + + + + k− ..... 20 1 21 P p( ) B P −1 =Thus () () n B P On p A = O ⇔ P p = −1 ( ) B On ⇔ p =Thus “ IF A and B ARE SIMILAR MATRICES THEN FOR ANY POLYNOMIAL p(λ); p (A) = On ⇔ p (B) = On ”. (8) Let A be any matrix. By A(A) we denote the set of all polynomials p(λ) such that  p(A) = On, i.e. A (A) = {p(λ) : p(A) = On} Now from (6) it follows that, “ IF A AND B ARE SIMILAR MATRICES THEN A(A) = A (B) ”.  The set A (A) is called the set of “ ANNIHILATING POLYNOMIALS OF A ”. Thus similar matrices have the same set of annihilating polynomials.  We shall discuss more about annihilating polynomials later. We now investigate the following question? Given an nxn matrix A, when is it similar to a “simple matrix”? What are simple matrices? The simplest matrix we know is the zero matrix On. Now A ∼ On ⇔ There is a nonsingular matrix P such that A = P-1On P = On. ∴ “ THE ONLY MATRIX SIMILAR TO On IS On ITSELF ”.  The next simple matrix we know is the identity matrix In. Now A ∼ In ⇔there is a nonsingular P such that A = P-1 In P ⇔ A = In.94 Thus “THE ONLY MATRIX SIMILAR TO In IS ITSELF ”. Similarly the only matrix similar to a scalar matrix kIn , (where k is a scalar),is kInitself. The next class of simple matrices are the DIAGONAL MATRICES. So we now ask the question “ Which type of nxn matrices are similar to diagonal matrices”?  Suppose now A is an nxn matrix; and A is similar to a diagonal matrix, D=12nλλλ        O (λI not necessarily distinct). Then there exists a nonsingular matrix P such that  P-1 A P = D ⇒ AP = PD &hellip;&hellip;&hellip;..(1) 11 12 1 121 22 2 21 2..... .......... .......... .....i ni nn n ni nnpp p ppp p p LetPpp p p   = M MMMM M=n n nnnna a aa a aa a aA.......... ..... ..... ...............1 221 22 211 12 112iiinipp LetPp     =     M denote the ith column of P.Now the ith column of AP is 11 1 12 2 121 1 22 2 211 2 2...........................................................i i n nii i n nin i n i nn niap ap apap ap apap ap ap  + ++  + ++  + ++which is equal to APi. Thus the ith column of AP, the l.h.s. of (1), is APi. Now the ith column of P D is 1 12 2ii iii ii iini i nip pp p Pp pλλ λ λλ        = =        M MThus the ith column of P D, the r.h.s. of (1), is λi Pi. Since l.h.s. = r.h.s. by (1) we have APi = λi Pi ; i = 1, 2, &hellip;., n &hellip;&hellip;&hellip;&hellip;&hellip;..(2) Note that since P is nonsingular no column of P can be zero vector. Thus none of the column vectors Pi are zero. Thus we conclude that, “IF A IS SIMILAR TO A DIAGONAL MATRIX D THEN THE DIAGONAL ENTRIES OF D MUST BE THE EIGENVALUES OF A AND IF P-1AP = D THEN THE ith COLUMN VECTOR OF P MUST BE AN EIGENVECTOR CORRESPON DING TO THE EIGENVALUE WHICH IS THE ith DIAGONAL ENTRY OF D”. Note:The n columns of P must be linearly independent since P is nonsingular and thus these n columns give us n linearly independent eigenvectors of A  Thus the above result can be restated as follows: A is similar to a diagonal matrix D and P-1 A P = D ⇒ A has n linearly independent eigenvectors; taking these as the columns of P we get P-1 A P we get D where the ith diagonal entry of D is the eigenvalue corresponding to the ith eigenvector, namely the Ithcolumn vector of P. . Conversely, it is now obvious that if A has n linearly independent eigenvectors then A is similar to a diagonal matrix D and if P is the matrix whose ith column is the eigenvector, then D is P-1 A P and ith diagonal entry of D is the eigenvalue corresponding to the ith eigenvector.  When does then a matrix have n linearly independent eigenvectors? It can be shown that a matrix A has n linearly independent eigenvectors ⇔ the algebraic multiplicity of each eigenvalue of A is equal to its geometric multiplicity. Thus A IS SIMILAR TO A DIAGONAL MATRIX⇔FOR EVERY EIGENVALUE OF A, ITS ALGEBRAIC MULTIPLICITY IS EQUAL TO ITS GEOMETRIC MULTPLICITY”.   RECALL; if () ( )( ) ( )akka a C λ = λ − λ λ − λ ..... λ − λ 1 21 2 where λ 1, λ 2, &hellip;.., λ k are the distinct eigenvalues of A, then ai is called the algebraic multiplicity of the eigenvalue λ i. Further, let { } x Ax x ωi = = λi :be the eigensubspace corresponding to λ i. Then gi = dim ωi is called the geometric multiplicity of λ i.  Therefore, we have, “ If A is an nxn matrix with 11 () ( ) ( ) k a a C λ λλ λλ =− − L k where λ 1, &hellip;.., λ k are the district eigenvalues of A, then A is similar to a diagonal matrix ⇔ ai = gi (=dimωi) ; 1≤ i ≤ k”. Example:Let us now consider −−−=16 8 78 3 49 4 4AOn page 87, we found the characteristic polynomial of A as C( λ ) = ( λ +1)2 ( λ - 3) Thus λ 1 = -1 ; a1 = 2 λ 2 = 3 ; a2 = 1On pages 83 and 84 we found, W1 = eigensubspace corresponding to λ = -1 += =201021: A1 A2 x xW2 = eigensubspace corresponding to λ = 3 = =211x : x kThus dim W1 = 2 ∴ g1 = 2  dim W2 = 1 ∴ g2 = 1 Thus a1 = 2 = g1 and hence A must be similar.  a2 = 1 = g2to a diagonal matrix. How do we get P such that P-1AP is a diagonal matrix? Recall the columns of P must be linearly independent eigenvectors. From ω1 we get two linearly eigenvectors, namely, 021 and 201 ; and from ω2 we get third as 211. Thus if we take these as columns and write, =0 2 22 0 11 1 1PThen −− −−= −2 1 12 2 1 12 1 0 11 P ; and it can be verified that −−−−− −−= −0 2 22 0 11 1 116 8 78 3 49 4 42 1 12 2 1 12 1 0 11 P AP −−=0 0 30 1 01 0 0 a diagonal matrix., whose diagonal entries are the eigenvalues of the matrix A. Thus we can conclude that A is similar to a diagonal matrix, i.e., P-1 AP = D ⇒ A has n linearly independent eigenvectors namely the n columns of P. Conversely, A has n linearly independent eigenvectors. ⇒ P-1 AP is a diagonal matrix where the columns of P are taken to be the n linearly eigenvectors.  We shall now see a class of matrices for which it is easy to decide whether they are similar to a diagonal matrix; and in which case the P-1 is easy to compute. But we shall first introduce some preliminaries.  If =n xxxx M21 ; =n yyyy M21 are any two vectors in Cn, we define the INNER PRODUCT OF x with y (which is denoted by (x,y)) as, ( ) n n x y = x y + x y + K + x y 1 1 2 2 , ∑==nii i x y1Example 1: If −= +12 iix ; = −iy 1 i1; then, ( x , y ) = i.1 + ( )( ) 2 + i 1 − i + ( ) − 1 (i) = i + ( )( ) ( )( ) 2 + i 1 + i + − 1 − i = 1 + 5 iWhereas ( y , x ) = 1 (i)+ ( )( ) 1 − i 2 + i + ( )i (− 1 ) = 1 − 5 iWe now observe some of the properties of the inner product, below: (1) For any vector x in Cn, we have ( ) , , 121∑ ∑ = == =niinii i x x x x xwhich is real ≥ 0. Further, ( ) ∑== ⇔ =nii x x x12, 0 0⇔ x i = 0;1 ≤ i ≤ nn ⇔ x = θ Thus,  (x,x) is real and ≥ 0 and (x,x)= 0 ⇔ x = θn   (2) ( )  = ∑ = ∑ = =nii inii i x y x y y x1 1, = ( ) y , x Thus,  ( )( ) x , y = y , x(3) For any complex number α, we have, ( ) () ∑ ∑ = == =nii inii i x y x y x y1 1α , α α = α ( ) x, yThus  (αx,y) = α (x,y) for any complex number α.  We note, ( )( ) x,αy = αy , x by (2)  = α ( ) y , x = α ( ) y , x = α ( ) x, y (4) ( )( ) ∑=+ = +nii i i x y z x y z1, ∑ ∑ = == +ninii i i i x z y z1 1 = ( )( ) x, y + x,zThus  (x + y, z) = (x,z) + (y,z) and similarly  (x, y + z) = (x, y) + (x, z) We say that two vectors x and y are ORTHOGONAL if (x, y) = 0. Example (1) If x = ,01;1−=−y iiithen, ( ) () x, y = 1 −1 + i( )i + ( )( ) − i 0 = -1 + 1 = 0 Thus x and y are orthogonal. (2) If −=−=11,1y aix ithen ( ) x, y = −1+ ai − i∴ x, y orthogonal  ( )( )a ii i iiiai ai⇔ = + = − + = −   +⇔ =⇔ − + + =11 1 11 0

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