[Journal of mathematics at home University] Question of 065th Nankai University's 2011 advanced algebra Postgraduate Entrance Exam

1 ($ 20' $) set $ {\ BF a} $ to the $ N $ square matrix with a rank of $1 $, $ {\ BF a} $ trace $ \ tr ({\ BF a}) = A \ NEQ 0 $. try to find all the feature values of $ {\ BF a} $ (write duplicates ).

A: Any two rows of $ \ rank ({\ BF a}) = 1 $ Zhi $ {\ BF a} $ are linearly related, $ \ Bex {\ bf a }=\ sex {\ BA {CCCC} B _1 & B _1 & \ cdots & B _n \ c_2b_1 & c_2b_2 & \ cdots & c_2b_n \\\ vdots & \ ddots & \ vdots \ c_nb_1 & c_nb_2 & \ cdots & c_nb_n \ EA }=\ sex {\ BA {CCC} C_1 \ vdots \ C_n \ EA} \ sex {\ BA {CCC} B _1 & \ cdots & B _n \ EA }, \ quad \ sex {c_1 = 1 \ atop \ MAX \ sev {B _ I}> 0 }. \ EEx $ you \ tr ({\ BF a}) = A \ NEQ 0 $, $ \ Bex \ sum _ {I = 1} ^ n c_ib_ I = A \ NEQ 0. \ EEx $ notice

(1) On the one hand, $ \ Bex {\ BF a} \ sex {\ BA {CCC} C_1 \ vdots \ C_n \ EA} & =&\ sex {\ BA {CCC} C_1 \ vdots \ C_n \ EA} \ sex {\ BA {CCC} B _1 & \ cdots & B _n \ EA} \ sex {\ BA {CCC} C_1 \ vdots \ C_n \ EA }\\\&\sex {\ sum _ {I = 1} ^ nb_ic_ I} \ sex {\ BA {CCC} C_1 \ vdots \\ c_n \ EA} = A \ sex {\ BA {CCC} C_1 \ vdots \ C_n \ EA }, \ EEx $ we know that $ A $ is the unique feature value of $ {\ BF a} $.

(2) On the other hand, $ \ Bex \ rank () = 1 & \ rA & {\ BF a} {\ BF x} = {\ BF 0} \ mbox {has} n-1 \ mbox {linear independent solutions }\\& \ rA & 0 \ mbox {Yes} {\ BF a} \ mbox {} n-1 \ mbox {HEAVY feature value }. \ EEx $ In summary, $ {\ BF a} $ n-1 $ heavy feature value $0 $, single heavy feature value $ A $.

 

2 ($ 20' $) set $ V $ to $4 $ Dimension Linear Space. $ \ ve_1, \ ve_2, \ ve_3, \ ve_4 $ is a group of bases, known Linear Transformations on $ V $ \ scrt $ on base $ \ ve_1, \ ve_2, \ ve_3, the matrix under \ ve_4 $ is $ \ Bex {\ BF t }=\ sex {\ BA {CCCC} 0 & 0 &-1 &-1 \ 0 & 1 & 2 & 2 \ 0 &-1 &-1 & 0 \ 0 & 0 & 0 & 1 \ EA }. \ EEx $

(1) obtain the feature values and feature vectors of $ \ scrt $;

(2) obtain the bases and dimensions of $ \ scrt $ and $ \ scrt $ respectively.

Answer:

(1) by $ \ Bex 0 & =& \ det (\ Lambda {\ BF e}-{\ BF t }) \\&=& \ det \ sex {\ BA {CCCC} \ Lambda & 0 & 1 & 1 \ 0 & \ Lambda-1 &-2 &-2 \ 0 & 1 & \ Lambda + 1 & 0 \ 0 & 0 & \ Lambda-1 \ EA} \ & = & (\ Lambda-1) \ Lambda (\ Lambda-I) (\ Lambda + I) \ EEx $ Zhi $ {\ BF t} $ (from but $ \ scrt $) the feature value is $ \ Bex \ lambda_1 = 1, \ quad \ lambda_2 = 0, \ quad \ lambda_3 = I, \ quad \ lambda_4 =-I. \ EEx $ again by $ \ Bex \ lambda_1 {\ BF e}-{\ BF t} = \ sex {\ BA {CCCC} 1 & 0 & 1 & 1 \ \ 0 & 0 &-2 &-2 \ 0 & 1 & 2 & 0 \ 0 & 0 & 0 \ EA} \ RRA \ sex {\ BA {CCCC} 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 \ 0 & 1 & 2 & 0 \ 0 & 0 & 0 & 0 \ EA }, \ EEx $ \ Bex \ lambda_2 {\ BF e}-{\ BF t }=\ sex {\ BA {CCCC} 0 & 0 & 1 & 1 \ 0 &-1 &-2 &-2 \ 0 & 1 & 0 \ 0 & 0 &-1 \ EA} \ RRA \ sex {\ BA {CCCC} 0 & 0 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ EA }, \ EEx $ \ Bex \ lambda_3 {\ BF e}-{\ BF t }=\ sex {\ BA {CCCC} I & 0 & 1 & 1 \ 0 & I-1 &-2 &-2 \ 0 & I + 1 & 0 \ 0 & 0 & i-1 \ EA} \ RRA \ sex {\ BA {CCCC} I & 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 1 & I + 1 & 0 \ 0 & 0 & 0 & 1 \ EA}, \ EEx $ \ Bex \ lambda_4 {\ BF e}-{\ BF t }=\ sex {\ BA {CCCC}-I & 0 & 1 & 1 \\ 0 &-i-1 &-2 &-2 \ 0 & 1 &-I + 1 & 0 \ 0 & 0 &-i-1 \ EA} \ RRA \ sex {\ BA {CCCC}-I & 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 1 &-I + 1 & 0 \ 0 & 0 & 0 & 0 & 1 \ EA} \ EEx $ Zhi $ {\ BF t} $ (and therefore $ \ scrt $) corresponding to $ \ lambda_1 $, $ \ lambda_2 $, $ \ lambda_3 $, $ \ lambda_4 $ feature vectors are $ \ Bex {\ BF \ ETA} _ 1 =\sex {\ BA {CCCC} 0 \ 2 \-1 \\ 1 \ EA }, \ quad {\ BF \ ETA} _ 2 = \ sex {\ BA {CCCC} 1 \ 0 \ 0 \ 0 \ EA }, \ quad {\ BF \ ETA} _ 3 = \ sex {\ BA {CCCC} I \-i-1 \ 1 \ 0 \ EA }, \ quad {\ BF \ ETA} _ 4 = \ sex {\ BA {CCCC}-I \ i-1 \ 1 \ 0 \ EA }. \ EEx $

(2) by $ \ Bex {\ BF t }=\ sex {\ BA {CCCC} 0 & 0 &-1 &-1 \ 0 & 1 & 2 & 2 \ 0 &-1 &-1 & 0 \ 0 & 0 & 0 & 1 \ EA} \ RRA \ sex {\ BA {CCCC} 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ EA} \ EEx $ \ rank ({\ BF t }) = 3 $, and $ \ Bex \ dim \ sex {\ Ker \ scrt} = 1, \ quad \ dim \ sex {\ im \ scrt} = 3. \ EEx $ a group of bases for $ \ Ker \ scrt $, which is the first small question, yi zhi $ {\ BF \ ETA} _ 2 $; $ {\ BF \ ETA} _ 1 $, $ {\ BF \ ETA} _ 3 $, $ {\ BF \ ETA} _ 4 $ is a group of base \ footnote {$ \ DPS {\ rank \ sex {{\ BF \ ETA} _ for $ \ im \ scrt $ }_ 1, {\ BF \ ETA} _ 3, {\ BF \ ETA} _ 4} = 3} $ .}.

 

3 ($ 20' $) set the real matrix $ \ DPS {\ bf a }=\ sex {\ BA {CCCC} 1 & 2 & 1 & 1 \ 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 \ EA} $, write $ {\ BF a} $ as the product of an orthogonal matrix $ {\ bf q} $ and an upper Triangle Matrix $ {\ BF t} $.

Answer: set \ footnote {the existence of QR decomposition is easily obtained by Schmidt's orthogonal process .} $ \ Bex {\ bf a }={\ bf q} {\ BF t}, \ EEx $ where $ q $ is an orthogonal array, $ T $ indicates the upper triangle array. then $ \ Bex {\ BF a} ^ t {\ bf a }={\ BF t} ^ t {\ bf q} {\ BF t }={ \ BF t} ^ t {\ BF t }. \ EEx $ then $ \ Bex {\ BF t} ^ t {\ BF t} & = & \ sex {\ sum _ {k = 1} ^ 4 T _{ ki} t _ {kJ }}=\ sex {\ sum _ {1 \ Leq k \ Leq \ min \ sed {I, j }}t _ {Ki} t _ {kJ }}\\&{\ BF a} ^ t {\ bf a }=\ sex {\ BA {CCCC} 2 & 2 & 1 & 2 \ 2 & 5 & 3 & 3 \ 1 & 3 & 2 & 2 \ 2 & 3 & 2 & 4 \ EA }. \ EEx $ hence $ {\ BF t} $ can be \ footnote {not unique. if the $ {\ BF t} $ symbol is given at the diagonal corner, it is unique .} $ \ Bex {\ BF t }=\ sex {\ BA {CCCC} \ SQRT {2} & \ SQRT {2} & \ frac {\ SQRT {2 }}{ 2} & \ SQRT {2} \ 0 & \ SQRT {3} & \ frac {2 \ SQRT {3 }}{ 3} & \ frac {\ SQRT {3}} {3} \ 0 & 0 & \ frac {\ SQRT {6 }}{ 6} & \ frac {\ SQRT {6 }}{ 3} \ 0 & 0 & 0 & 1 \ EA }. \ EEx $ and $ \ Bex {\ bf q} & =& {\ bf a }{\ BF t }^{-1 }\\& = & \ sex {\ BA {CCCC} 1 & 2 & 1 & 1 \ 1 & 0 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \ EA} \ sex {\ BA {CCCC} \ frac {\ SQRT {2 }}{ 2} &-\ frac {\ SQRT {3 }}{ 3} & \ frac {\ SQRT {6 }}{ 6} &-1 \ 0 & \ frac {\ SQRT {3 }}{ 3} &-\ frac {2 \ SQRT {6 }} {3} & 1 \ 0 & 0 & \ SQRT {6} &-2 \ 0 & 0 & 0 & 1 \ EA} \ & \ sex {\ sex {\ BF t }, {\ BF e }}\ RRA \ sex {\ BF e }, {\ BF t} ^ {-1 }}\\\& = & \ sex {\ BA {CCCC} \ frac {\ SQRT {2 }}{ 2} & \ frac {\ SQRT {3 }}{ 3} &-\ frac {\ SQRT {6 }}{ 6} & 0 \ frac {\ SQRT {2 }}{ 2} &-\ frac {\ SQRT {3 }}{ 3} & \ frac {\ SQRT {6 }}{ 6} & 0 \ 0 & \ frac {\ SQRT {3 }}{ 3} & \ frac {\ SQRT {6 }}{ 3} & 0 \ 0 & 0 & 1 \ EA }. \ EEx $

 

4 ($ 20' $): Set $ {\ BF a} $ as the antisymmetric matrix. proof: $ {\ BF e}-{\ BF a} ^ {10} $ must be a positive matrix.

Proof: we prove the more general conclusion: $ \ Bex {\ BF e}-{\ BF a} ^ {2 (2 k + 1 )} \ mbox {positive definite matrix} \ quad \ sex {k =, 2, \ cdots }. \ EEx $ notice $ \ Bex {\ BF a} ^ t = {\ BF a} \ Ra \ left \ {\ BA {ll} {\ BF a} ^ {t2l} = {\ BF a} ^ {2L} \ {\ BF a} ^ {T (2L + 1 )} =-{\ BF a} ^ {2L + 1} \ EA \ right. \ quad \ sex {L = 1, 2, 3, \ cdots }. \ EEx $ so $ {\ BF e}-{\ BF a} ^ {2 (2 k + 1)} $ is a symmetric array, for $ \ forall \ {\ BF x} \ In \ BBR ^ N $, $ \ Bex {\ BF x} ^ t \ sex {\ BF e}-{\ BF a} ^ {2 (2 k + 1 )}} {\ BF x} <=& {\ BF x} ^ t {\ BF x}-{\ BF x} ^ t {\ BF a} ^ {2 k + 1} \ cdot {\ BF a} ^ {2 k + 1} {\ BF x }\\\& =&{\ BF x} ^ t {\ BF x }+ {\ BF x} ^ t \ sex {\ BF a} ^ t} ^ {2 k + 1} \ cdot {\ BF a} ^ {2 k + 1} {\ BF x} \ & = & {\ BF x} ^ t {\ BF x} + {\ BF x} ^ t \ sex {\ BF a} ^ {2 k + 1} ^ t \ cdot {\ BF a} ^ {2 k + 1} {\ BF x} \ quad \ sex {(BC) ^ t = C ^ TB ^ t }\\\&{\ BF x} ^ t {\ BF x }+ \ sex {{\ bf a }^{ 2 K + 1} {\ BF x }}^ t \ sex {\ BF a} ^ {2 k + 1} {\ BF x }}\\\& = & \ sev {{ \ BF x }}^ 2 + \ sev {\ BF a} ^ {2 k + 1} {\ BF x }}^ 2 \ & \ geq & \ sev {{\ BF x }}^ 2. \ EEx $ and $ {\ BF e}-{\ BF a} ^ {2 (2 k + 1)} $ is a definite array.

 

5 ($ 15' $) is set to $ V $, which is an Euclidean space. $ \ scrt $ is a ing between $ V $ and $ V $, meeting the following conditions: $ \ Bex \ sev {\ scrt {\ BF \ Alpha }}=\ sev {\ BF \ Alpha }}, \ qquad \ forall \ {\ BF \ Alpha} \ In v. \ EEx $ question: $ \ scrt $ must be an orthogonal transformation on $ V $? Reasons.

A: $ \ scrt $ is not necessarily an orthogonal transformation on $ V $. because $ \ scrt $ is not necessarily a linear transformation. for example, for $ v = \ BBR ^ 2 $, get $ \ Bex \ scrt \ sex {\ BA {CC} X \ Y \ EA }=\ left \ {\ BA {ll} \ sex {\ BA {CC} x-Y \ SQRT {2xy} \ EA }, & XY \ geq 0, \ sex {\ BA {CC} X + Y \ SQRT {-2xy} \ EA}, & XY <0, \ EA \ right. \ EEx $ is suitable for $ \ Bex \ sev {\ scrt \ sex {\ BA {CC} X \ Y \ EA} =\ sev {\ sex {\ Ba {CC} X \ Y \ EA }}, \ qquad \ forall \ sex {\ BA {CC} X \ Y \ EA} \ In \ BBR ^ 2. \ EEx $ but $ \ Bex \ scrt \ sex {\ BA {CC} 1 \ 1 \ EA} = \ sex {\ BA {CC} 0 \ SQRT {2} \ EA} \ NEQ \ sex {\ BA {CC} 0 \ 0 \ EA }=\ scrt \ sex {\ BA {CC} 1 \ 0 \ EA} + \ scrt \ sex {\ BA {CC} 0 \ 1 \ EA }. \ EEx $

 

6 ($ 15' $) set $ {\ bf a }$, $ {\ BF B} $ to the square matrix of $ N $ on the number field $ \ BBP $, satisfies equation $ A {\ BF a} ^ 2 + B {\ BF a} {\ BF B} + c {\ bf B }={\ BF 0 }$, where $ A, B, and C $ are non-zero constants. proof: $ \ sex {c {\ BF e} + {\ bf a }}$ is a reversible matrix.

Proof: Use the reverse verification method. if $ \ sex {c {\ BF e} + {\ BF a} $ is irreversible, then $ \ Bex & \ det \ sex {c {\ BF e} + B {\ BF A }}= 0 \ & \ rA & \ det \ sex {C {\ BF e} + B {\ BF a} ^ t = 0 \ & \ rA & \ exists \ x_0 \ NEQ 0, \ s. t. \ sex {c {\ BF e} + B {\ BF a} ^ t {\ BF x} _ 0 = {\ BF 0} \ & \ rA &{ \ BF x }_0 ^ t \ sex {c {\ BF e} + B {\ BF A }}={\ BF 0 }\\& \ rA & {\ BF x} _ 0 ^ t {\ BF a} =-\ frac {c} {B} {\ BF x} _ 0 ^ t \ quad \ sex {\ mbox {so-called left feature vector }}. \ EEx $ and $ A {\ BF a} ^ 2 + B {\ BF a} {\ BF B} + c {\ bf B }={\ BF 0} $ $ \ Bex {\ BF 0} & =& {\ BF x} _ 0 ^ t \ sex {B {\ BF a} + c {\ BF e }}{ \ bf B }\\& = &-a {\ BF x} _ 0 ^ t {\ BF a} ^ 2 \\& = & A \ cdot \ frac {c} {B} {\ BF x }_0 ^ t {\ bf a }\\\& = &-A \ frac {C ^ 2} {B ^ 2} {\ BF x} _ 0 ^ t. \ EEx $ this is a contradiction \ footnote {$ A, B, C $ non-zero, $ {\ BF x }_0 \ NEQ {\ BF 0 }$ .}.

 

7 ($ 15' $) set $ {\ bf a }$, $ {\ BF B} $ to the square matrix of $ N $ on the number field $ \ BBP $, and $ \ rank ({\ BF a}) =\ rank ({\ BF B} {\ BF a}) $. proof: For any natural number $ L $, $ \ rank ({\ BF a} ^ L) = \ rank ({\ BF B} {\ BF a} ^ L) $.

Proof:

(1) first prove $ \ bee \ label {nk11gd: 7: eq} \ rank ({\ BF }) =\ rank ({\ bf B }{\ BF }) \ LRA {\ BF a} {\ BF x} = {\ BF 0} \ mbox {And} {\ BF B} {\ BF a} {\ BF x} = {\ BF 0} \ mbox {has the same solution }. \ EEE $ \ la $ \ Bex \ rank ({\ BF a}) & = & N-\ dim \ sed {{\ BF x }; \{\ bf a }{\ BF x }={\ BF 0 }\\\& = & N-\ dim \ sex {{\ BF x }; \ {\ BF B} {\ BF a} {\ BF x} = {\ BF 0 }\\\& = & \ rank ({\ BF B} {\ BF }). \ EEx $ \ rA $ obviously there are $ \ Bex \ sed {{\ BF x }; \ {\ BF a} {\ BF x }={\ BF 0 }}\ subset \ sed {\ BF x }; \ {\ BF B} {\ BF a} {\ BF x} = {\ BF 0 }}. \ EEx $ you $ \ Bex \ dim \ sed {\ BF x }; \ {\ bf a }{\ BF x }={\ BF 0 }}& = & N-\ rank ({\ BF }) \\& = & N-\ rank \ sex {{\ bf B }{\ BF A }\\\& = & \ dim \ sex {{\ BF x }; \{\ bf B }{\ BF A }{\ BF x }={\ BF 0 }}, \ EEx $ we know that \ footnote {linear subspaces with the same dimension as finite-dimensional linear spaces are themselves .} $ \ Bex \ sed {{\ BF x }; \{\ bf a }{\ BF x }={\ BF 0 }}=\ sex {{\ BF x }; \ {\ bf B }{\ BF A }{\ BF x }={\ BF 0 }}\ EEx $

(2) prove again: $ \ rank ({\ BF a} ^ L) = \ rank ({\ BF B} {\ BF a} ^ L) $, $ \ forall \ L = 1, 2, 3, \ cdots $. if $ L = 1 $, it is obvious. if $ L \ geq 2 $, press \ eqref {nk11gd: 7: eq} to verify: $ \ Bex {\ BF B} {\ BF a} ^ L {\ BF x }={\ BF 0} & \ rA & {\ BF B} {\ BF} \ sex {\ BF a} ^ {L-1 {\ BF x }={\ BF 0 }\\ \ rA & {\ BF a} \ sex {{ \ BF a} ^ l-1} {\ BF x }={ \ BF 0} \ quad \ sex {\ rank ({\ BF }) =\ rank ({\ BF B} {\ BF a}) \ mbox {And} \ eqref {nk11gd: 7: EQ }}\\ \ rA & {\ BF a} ^ L {\ BF x }={\ BF 0 }. \ EEx $

 

8 ($ 15' $) set $ V $ to $ 4N $ dimension linear space in the complex field. proof: There is a linear transformation on $ V $ \ scrt $ to make $ \ scrt ^ 4 =-Id $, where $ ID $ is a constant transformation. it is proved that the matrix of a linear transformation meeting the above conditions must be a diagonal matrix.

Proof: Remember $ \ Bex {\ BF e} _ I = (\ underbrace {0, \ cdots, 1 }_{ I \ mbox {}}, 0, \ cdots), \ quad \ sex {1 \ Leq I \ Leq 4N }. \ EEx $ then $ \ sed {\ BF e} _ I }_{ I = 1} ^ {4N} $ is a group of bases for $ V $. take the linear transformation on $ V $ \ scrt $ to make $ \ Bex \ scrt ({\ BF e} _ I) =\ frac {\ SQRT {2 }}{ 2} (1 + I) {\ BF e} _ I, \ EEx $ then $ \ Bex \ scrt ^ 4 \ sex {\ BF e }_ I }=\ SEZ {\ frac {\ SQRT {2 }}{ 2} (1 + I )} ^ 4 {\ BF e} _ I =-ID \ sex {\ BF e} _ I }. \ EEx $

 

9 ($ 10' $) A $ N $ square matrix $ {\ BF a} $ on the number field $ \ BBP $ is called a power zero, if there is a natural number $ M $ to make $ {\ BF a} ^ m = {\ BF 0} $. set $ {\ bf a }=\ sex {A _ {IJ }}_{ n \ times n} $ to a zero-power matrix, and $ \ Bex a _ {12} \ NEQ 0, \ quad A _ {13} = 0, \ quad A _ {22} = 0, \ quad A _ {24} \ NEQ 0. \ EEx $ proof: The matrix does not exist $ {\ BF B} $ make $ {\ BF B} ^ {n-1 }={ \ BF a} $.

Proof: Use the reverse verification method. if $ \ Bex \ exists \ {\ BF B} \ In \ BBP ^ {n \ times n}, \ s. t. \ {\ BF B} ^ {n-1 }={\ BF }. \ EEx $ then $ \ Bex {\ BF a} ^ m = {\ BF B} ^ {(n-1) m }={ \ BF 0 }. \ EEx $ by \ footnote {By \ eqref {nk11gd: 7: eq }, only $ \ Bex {\ BF B} ^ K {\ BF x }={\ BF 0} \ LRA {\ BF B} ^ {k + 1 }{\ BF x }={\ BF 0 }\ quad \ sex {k \ geq n }. \ EEx $ \ rA $ apparently. $ \ la $ use reverse identification. if $ {\ BF B} ^ K {\ BF x} \ NEQ 0 $, then $ {\ BF x}, \ {\ BF B} {\ BF x }, \ {\ BF B} ^ K {\ BF x} $ this $ k + 1 (\ geq n + 1) $ Vector Linear Independence: $ \ Bex \ sum _ {I = 0} ^ K C_ I {\ BF B} ^ I {\ BF x} = {\ BF 0} & \ rA & c_0 = 0 \ quad \ sex {\ mbox {use} {\ BF B} ^ k \ mbox {act on both ends} \ & \ rA & c_1 = 0 \ quad \ sex {\ mbox {with} {\ BF B} ^ {k-1} \ mbox {acting on both ends} \ & \ rA & \ cdots \ & \ rA & C _ {k-1} = 0 \ quad \ sex {\ mbox {use} {\ BF B} \ mbox {act on both ends} \ & \ rA & c_k = 0. \ EEx $ this is a conflict .} $ \ Bex \ rank ({\ BF B} ^ n) = \ rank ({\ BF B} ^ K) \ quad \ sex {k \ geq n} \ EEx $ \ Bex \ rank ({\ BF B} ^ N) = \ rank ({\ BF B} ^ {(n-1) m}) = 0, \ EEx $ \ Bex {\ BF B} ^ n = {\ BF 0 }. \ EEx $ and the feature values of $ {\ BF B} $ are all $0 $ \ footnote {$ \ Bex \ left. \ BA {RR} {\ BF B} {\ BF x }=\ Lambda {\ BF x }\\{ \ BF x} \ NEQ {\ BF 0} \ EA \ right \} \ Ra \ left. \ BA {RR} {\ BF 0 }={\ BF B} ^ n {\ BF x }=\ Lambda ^ n {\ BF x }\{\ BF x} \ NEQ {\ BF 0} \ EA \ right \} \ Ra \ Lambda = 0. \ EEx $ }. note that $ \ Bex {\ BF B} ^ {n-1} = {\ BF a} \ NEQ 0 \ quad \ sex {\ mbox {minimum} A _ {12 }\ NEQ 0, \ A _ {23} \ NEQ 0}, \ EEx $ our $ {\ BF B} $ Jordan standard type is \ footnote {according to Jordan standard type theory, reversible array $ {\ BF t} $, make $ {\ BF t} ^ {-1} {\ BF B} {\ BF t }={ \ BF J} $, $ {\ bf j} $ indicates some Jordan blocks. if $1 $ in $ {\ bf j} $ is less than or equal to $ N-2 $, $ \ Bex {\ BF t} ^ {-1} {\ BF B} ^ {L + 1} {\ BF t }={\ BF J} ^ {L + 1 }={\ BF 0 }\ Ra {\ bf B }^{ L + 1 }={\ BF 0 }\ quad \ sex {L + 1 \ leq n-1 }. \ EEx $ conflicts with $ {\ BF B} ^ {n-1 }={\ BF a} \ NEQ 0 $ .} $ \ Bex {\ bf j }=\ sex {\ BA {CCCCC} 0 & 1 & \\\ & 0 & \ ddots & \\& \ ddots & \ ddots & \\& & 0 & 1 \\& & 0 \ EA }. \ EEx $ so, $ \ Bex \ rank ({\ BF B}) = \ rank ({\ BF J}) = n-1. \ EEx $ but $ \ Bex & {\ BF a} {\ bf B }={\ BF B} ^ n ={\ BF 0 }\\ & \ rA & \ rank ({\ BF }) + \ rank ({\ BF B }) \ Leq n \ quad \ sex {\ mbox {linear equations theory }\\\& \ rA & \ rank ({\ BF a}) \ Leq 1. \ EEx $ this corresponds to $ \ Bex a _ {12} \ NEQ 0, \ A _ {13} = 0, \ A _ {22} = 0, \ A _ {24} \ NEQ 0 \ Ra \ rank ({\ BF a}) \ geq 2 \ EEx $ conflict. now, the question is verified.

 

 

Feelings

 

There are so many ''mathematical analytics ''in Higher Algebra. It is good to calculate or give an example;

There are also so many ''proof method'' in Higher Algebra, which leads to conflicts with existing knowledge or questions.